3. Using the definition of integration, find \(y\) in terms of \(x\).
3(i) \(\dfrac{dy}{dx}=3x^2+2\)
Given \(\dfrac{dy}{dx}=3x^2+2\). Integrate both sides:
\[
y=\int(3x^2+2)\,dx = 3\int x^2dx + 2\int dx
= 3\cdot\frac{x^3}{3} + 2x + C = x^3 + 2x + C.
\]
\[
y=\int(3x^2+2)\,dx = 3\int x^2dx + 2\int dx
= 3\cdot\frac{x^3}{3} + 2x + C = x^3 + 2x + C.
\]
\(\boxed{y=x^3+2x+C}\)
3(ii) \(dy=(ax+b)dx\)
Integrate:
\[
y=\int (ax+b)dx = a\int xdx + b\int dx
= \tfrac{a}{2}x^2 + bx + C.
\]
\[
y=\int (ax+b)dx = a\int xdx + b\int dx
= \tfrac{a}{2}x^2 + bx + C.
\]
\(\boxed{y=\tfrac{a}{2}x^2+bx+C}\)
3(iii) \(\dfrac{dy}{dx}=2\sin2x\)
Integrate:
\[
y=\int 2\sin2x\,dx = 2\left(-\tfrac{1}{2}\cos2x\right) + C = -\cos2x + C.
\]
\[
y=\int 2\sin2x\,dx = 2\left(-\tfrac{1}{2}\cos2x\right) + C = -\cos2x + C.
\]
\(\boxed{y=-\cos2x+C}\)
3(iv) \(\dfrac{dy}{dx}=\sqrt{x},\; \text{given } y(4)=3\)
Write \(\sqrt{x}=x^{1/2}\). Integrate:
\[
y=\int x^{1/2}dx = \tfrac{2}{3}x^{3/2} + C.
\]
Use condition \(y(4)=3\):
\[
3 = \tfrac{2}{3}\cdot 4^{3/2} + C = \tfrac{16}{3} + C \Rightarrow C = -\tfrac{7}{3}.
\]
So
\[
y=\tfrac{2}{3}x^{3/2} – \tfrac{7}{3}.
\]
\[
y=\int x^{1/2}dx = \tfrac{2}{3}x^{3/2} + C.
\]
Use condition \(y(4)=3\):
\[
3 = \tfrac{2}{3}\cdot 4^{3/2} + C = \tfrac{16}{3} + C \Rightarrow C = -\tfrac{7}{3}.
\]
So
\[
y=\tfrac{2}{3}x^{3/2} – \tfrac{7}{3}.
\]
\(\boxed{y=\tfrac{2}{3}x^{3/2}-\tfrac{7}{3}}\)
4. Integrate each function w.r.t. \(x\).
4(i) \(\displaystyle \int x^{2/3}\,dx\)
\(\int x^{n}dx = \dfrac{x^{n+1}}{n+1}+C\). So
\(\int x^{2/3}dx = \dfrac{x^{5/3}}{5/3} + C = \dfrac{3}{5}x^{5/3} + C.\)
\(\int x^{2/3}dx = \dfrac{x^{5/3}}{5/3} + C = \dfrac{3}{5}x^{5/3} + C.\)
\(\boxed{\tfrac{3}{5}x^{5/3}+C}\)
4(ii) \(\displaystyle \int \dfrac{2}{\sqrt[4]{x^3}}\,dx\)
\(\dfrac{2}{\sqrt[4]{x^3}} = 2x^{-3/4}\).
\[
\int 2x^{-3/4}dx = 2\cdot \frac{x^{1/4}}{1/4} + C = 8x^{1/4} + C = 8\sqrt[4]{x} + C.
\]
\[
\int 2x^{-3/4}dx = 2\cdot \frac{x^{1/4}}{1/4} + C = 8x^{1/4} + C = 8\sqrt[4]{x} + C.
\]
\(\boxed{8\sqrt[4]{x}+C}\)
4(iii) \(\displaystyle \int (x^4-2x^2+5)\,dx\)
Integrate termwise:
\[
\tfrac{x^5}{5} – \tfrac{2x^3}{3} + 5x + C.
\]
\[
\tfrac{x^5}{5} – \tfrac{2x^3}{3} + 5x + C.
\]
\(\boxed{\tfrac{x^5}{5}-\tfrac{2x^3}{3}+5x+C}\)
4(iv) \(\displaystyle \int\Big(x-\dfrac{2}{x}\Big)^3 dx\)
Expand:
\[
\Big(x-\tfrac{2}{x}\Big)^3 = x^3 -6x + \tfrac{12}{x} – \tfrac{8}{x^3}.
\]
Integrate:
\[
\tfrac{x^4}{4} – 3x^2 + 12\ln|x| + 4x^{-2} + C.
\]
\[
\Big(x-\tfrac{2}{x}\Big)^3 = x^3 -6x + \tfrac{12}{x} – \tfrac{8}{x^3}.
\]
Integrate:
\[
\tfrac{x^4}{4} – 3x^2 + 12\ln|x| + 4x^{-2} + C.
\]
\(\boxed{\tfrac{x^4}{4}-3x^2+12\ln|x|+\tfrac{4}{x^2}+C}\)
4(v) \(\displaystyle \int(x+3)(x^2-5)\,dx\)
Multiply: \((x+3)(x^2-5)=x^3+3x^2-5x-15\).
Integrate:
\[
\tfrac{x^4}{4}+x^3-\tfrac{5x^2}{2}-15x + C.
\]
Integrate:
\[
\tfrac{x^4}{4}+x^3-\tfrac{5x^2}{2}-15x + C.
\]
\(\boxed{\tfrac{x^4}{4}+x^3-\tfrac{5x^2}{2}-15x+C}\)
4(vi) \(\displaystyle \int \dfrac{1}{\sqrt{x}}\Big(\sqrt{x}+\dfrac{1}{\sqrt{x}}\Big)^2 dx\)
Expand: \((\sqrt{x}+\tfrac{1}{\sqrt{x}})^2 = x + 2 + \tfrac{1}{x}\).
Multiply by \(x^{-1/2}\): \(x^{1/2}+2x^{-1/2}+x^{-3/2}\).
Integrate:
\[
\tfrac{2}{3}x^{3/2} + 4\sqrt{x} – 2x^{-1/2} + C.
\]
Multiply by \(x^{-1/2}\): \(x^{1/2}+2x^{-1/2}+x^{-3/2}\).
Integrate:
\[
\tfrac{2}{3}x^{3/2} + 4\sqrt{x} – 2x^{-1/2} + C.
\]
\(\boxed{\tfrac{2}{3}x^{3/2}+4\sqrt{x}-\dfrac{2}{\sqrt{x}}+C}\)
5. Evaluate the following integrals.
5(i) \(\displaystyle \int e^{3\log x}\,dx\)
\(e^{3\ln x}=x^3\). So \(\int x^3 dx = \tfrac{x^4}{4} + C.\)
\(\boxed{\tfrac{x^4}{4}+C}\)
5(ii) \(\displaystyle \int \dfrac{e^{5x}-2e^{3x}+3}{e^x}\,dx\)
Divide: \(=e^{4x}-2e^{2x}+3e^{-x}\).
Integrate: \(\tfrac{e^{4x}}{4}-e^{2x}-3e^{-x}+C.\)
Integrate: \(\tfrac{e^{4x}}{4}-e^{2x}-3e^{-x}+C.\)
\(\boxed{\tfrac{e^{4x}}{4}-e^{2x}-3e^{-x}+C}\)
5(iii) \(\displaystyle \int \dfrac{e^{5x}+e^{3x}}{e^x+e^{-x}}\,dx\)
Simplify: \(\dfrac{e^{3x}(e^{2x}+1)}{e^{-x}(e^{2x}+1)}=e^{4x}.\)
So integral \(=\tfrac{e^{4x}}{4}+C.\)
So integral \(=\tfrac{e^{4x}}{4}+C.\)
\(\boxed{\tfrac{e^{4x}}{4}+C}\)
5(iv) \(\displaystyle \int (e^{2\log x}-2e^{-3\log x})\,dx\)
\(e^{2\ln x}=x^2,\; e^{-3\ln x}=x^{-3}.\) So integrand \(=x^2-2x^{-3}\).
Integrate: \(\tfrac{x^3}{3}+x^{-2}+C.\)
Integrate: \(\tfrac{x^3}{3}+x^{-2}+C.\)
\(\boxed{\tfrac{x^3}{3}+\dfrac{1}{x^2}+C}\)
5(v) \(\displaystyle \int e^{-4x}\,dx\)
\(\int e^{-4x}dx = -\tfrac{1}{4}e^{-4x} + C.\)
\(\boxed{-\tfrac{1}{4}e^{-4x}+C}\)
6. Integrate.
6(i) \(\displaystyle \int 5^{2x} dx\)
\(5^{2x}=e^{2x\ln5}\). So \(\int e^{2x\ln5}dx = \dfrac{5^{2x}}{2\ln5} + C.\)
\(\boxed{\dfrac{5^{2x}}{2\ln5}+C}\)
6(ii) \(\displaystyle \int (x^4+4^x)dx\)
\(\int x^4dx=\tfrac{x^5}{5},\; \int 4^x dx=\dfrac{4^x}{\ln4}.\)
So \(\tfrac{x^5}{5}+\dfrac{4^x}{\ln4}+C.\)
So \(\tfrac{x^5}{5}+\dfrac{4^x}{\ln4}+C.\)
\(\boxed{\tfrac{x^5}{5}+\dfrac{4^x}{\ln4}+C}\)
6(iii) \(\displaystyle \int (e^{x\log a}+e^{a\log x})dx\)
\(e^{x\log a}=a^x,\; e^{a\log x}=x^a\).
Integrate: \(\dfrac{a^x}{\ln a} + \dfrac{x^{a+1}}{a+1} + C.\)
Integrate: \(\dfrac{a^x}{\ln a} + \dfrac{x^{a+1}}{a+1} + C.\)
\(\boxed{\dfrac{a^x}{\ln a}+\dfrac{x^{a+1}}{a+1}+C}\)
7. Evaluate the following integrals (trigonometric).
7(i) \(\displaystyle \int \sin^2 2x\,dx\)
\(\sin^2 t = \tfrac{1-\cos2t}{2}\).
\(\sin^2 2x=\tfrac{1-\cos4x}{2}\).
\[
\int \sin^2 2x\,dx = \tfrac{x}{2} – \tfrac{\sin4x}{8} + C.
\]
\(\sin^2 2x=\tfrac{1-\cos4x}{2}\).
\[
\int \sin^2 2x\,dx = \tfrac{x}{2} – \tfrac{\sin4x}{8} + C.
\]
\(\boxed{\tfrac{x}{2}-\tfrac{\sin4x}{8}+C}\)
7(ii) \(\displaystyle \int \dfrac{2\tan x}{1+\tan^2x}\,dx\)
\(1+\tan^2x = \sec^2x\). So integrand \(= \dfrac{2\tan x}{\sec^2x} = 2\sin x\cos x = \sin2x\).
\[
\int \sin2x\,dx = -\tfrac{\cos2x}{2} + C.
\]
\[
\int \sin2x\,dx = -\tfrac{\cos2x}{2} + C.
\]
\(\boxed{-\tfrac{\cos2x}{2}+C}\)
7(iii) \(\displaystyle \int \sec x(\sec x+\tan x)\,dx\)
\(\sec x(\sec x+\tan x)=\sec^2x+\sec x\tan x\).
\[
\int = \tan x + \sec x + C.
\]
\[
\int = \tan x + \sec x + C.
\]
\(\boxed{\tan x + \sec x + C}\)
7(iv) \(\displaystyle \int \csc x(\csc x-\cot x)\,dx\)
\(\csc x(\csc x-\cot x)=\csc^2x-\csc x\cot x\).
\[
\int = -\cot x + \csc x + C.
\]
\[
\int = -\cot x + \csc x + C.
\]
\(\boxed{-\cot x+\csc x + C}\)
7(v) \(\displaystyle \int \cos x^\circ\,dx\)
Convert degrees: \(x^\circ = \dfrac{\pi x}{180}\). Let \(k=\dfrac{\pi}{180}\).
\[
\int \cos(kx)\,dx = \frac{\sin(kx)}{k} + C
\Rightarrow \frac{180}{\pi}\sin x^\circ + C.
\]
\[
\int \cos(kx)\,dx = \frac{\sin(kx)}{k} + C
\Rightarrow \frac{180}{\pi}\sin x^\circ + C.
\]
\(\boxed{\dfrac{180}{\pi}\sin x^\circ + C}\)
7(vi) \(\displaystyle \int \cot^2\theta\,d\theta\)
\(\cot^2\theta=\csc^2\theta-1\).
\[
\int = -\cot\theta – \theta + C.
\]
\[
\int = -\cot\theta – \theta + C.
\]
\(\boxed{-\cot\theta-\theta + C}\)
7(vii) \(\displaystyle \int \cos^2\theta\,d\theta\)
\(\cos^2\theta=\tfrac{1+\cos2\theta}{2}\).
\[
\int = \tfrac{\theta}{2} + \tfrac{\sin2\theta}{4} + C.
\]
\[
\int = \tfrac{\theta}{2} + \tfrac{\sin2\theta}{4} + C.
\]
\(\boxed{\tfrac{\theta}{2}+\tfrac{\sin2\theta}{4}+C}\)
7(viii) \(\displaystyle \int \dfrac{1+\tan^2x}{1+\cot^2x}\,dx\)
\(1+\tan^2x=\sec^2x,\;1+\cot^2x=\csc^2x\).
So integrand \(=\dfrac{\sec^2x}{\csc^2x}=\tan^2x\).
\[
\int \tan^2x\,dx = \tan x – x + C.
\]
So integrand \(=\dfrac{\sec^2x}{\csc^2x}=\tan^2x\).
\[
\int \tan^2x\,dx = \tan x – x + C.
\]
\(\boxed{\tan x – x + C}\)
7(ix) \(\displaystyle \int \dfrac{\sec x + 2\cot^2x + \cos^2x}{\cos x}\,dx\)
Simplify:
\[
\frac{\sec x + 2\cot^2x + \cos^2x}{\cos x}
= \sec^2x + 2\cot x\csc x + \cos x.
\]
Integrate termwise:
\[
\tan x – 2\csc x + \sin x + C.
\]
\[
\frac{\sec x + 2\cot^2x + \cos^2x}{\cos x}
= \sec^2x + 2\cot x\csc x + \cos x.
\]
Integrate termwise:
\[
\tan x – 2\csc x + \sin x + C.
\]
\(\boxed{\tan x – 2\csc x + \sin x + C}\)
7(x) \(\displaystyle \int \dfrac{\sec^2x}{\csc^2x}\,dx\)
\(\dfrac{\sec^2x}{\csc^2x}=\tan^2x\). So
\[
\int \tan^2x\,dx = \tan x – x + C.
\]
\[
\int \tan^2x\,dx = \tan x – x + C.
\]
\(\boxed{\tan x – x + C}\)
8. If \(\dfrac{dy}{dx}=2x^2-3\) at any point \((x,y)\) on a curve, find the family of curves.
8. Integrate
\[
y=\int(2x^2-3)dx = \tfrac{2}{3}x^3 – 3x + C.
\]
y=\int(2x^2-3)dx = \tfrac{2}{3}x^3 – 3x + C.
\]
\(\boxed{y=\tfrac{2}{3}x^3 – 3x + C}\)
1(i) \(\displaystyle \int \dfrac{3x^{2}-4x+5}{\sqrt{x}}\,dx\)
\(\dfrac{3x^{2}-4x+5}{\sqrt{x}} = 3x^{3/2}-4x^{1/2}+5x^{-1/2}\)
\(\Rightarrow \int 3x^{3/2}-4x^{1/2}+5x^{-1/2}\,dx = \frac{6}{5}x^{5/2}-\frac{8}{3}x^{3/2}+10x^{1/2}+C\)
\(\boxed{\frac{6}{5}x^{5/2}-\frac{8}{3}x^{3/2}+10x^{1/2}+C}\)
1(ii) \(\displaystyle \int \dfrac{(x^{2}+1)^{2}}{x^{3}}\,dx\)
\((x^{2}+1)^{2}=x^{4}+2x^{2}+1\Rightarrow\dfrac{x^{4}+2x^{2}+1}{x^{3}}=x+\dfrac{2}{x}+\dfrac{1}{x^{3}}\)
\(\Rightarrow \int (x+\dfrac{2}{x}+\dfrac{1}{x^{3}})\,dx = \frac{x^{2}}{2}+2\ln|x|-\frac{1}{2x^{2}}+C\)
\(\boxed{\frac{x^{2}}{2}+2\ln|x|-\frac{1}{2x^{2}}+C}\)
1(iii) \(\displaystyle \int (a^{2/3}+x^{2/3})^{3}\,dx\)
\((a^{2/3}+x^{2/3})^{3} = a^{2}+3a^{4/3}x^{2/3}+3a^{2/3}x^{4/3}+x^{2}\)
\(\Rightarrow \int(a^{2}+3a^{4/3}x^{2/3}+3a^{2/3}x^{4/3}+x^{2})\,dx\)
\(\ = a^{2}x + 3a^{4/3}\int x^{2/3}\,dx + 3a^{2/3}\int x^{4/3}\,dx + \int x^{2}\,dx\)
\(\Rightarrow a^{2}x + 3a^{4/3}\Big(\frac{3x^{5/3}}{5}\Big) + 3a^{2/3}\Big(\frac{3x^{7/3}}{7}\Big) + \frac{x^{3}}{3} + C\)
\(\Rightarrow \boxed{a^{2}x + \frac{9a^{4/3}x^{5/3}}{5} + \frac{9a^{2/3}x^{7/3}}{7} + \frac{x^{3}}{3} + C}\)
\(\boxed{a^{2}x + \frac{9a^{4/3}x^{5/3}}{5} + \frac{9a^{2/3}x^{7/3}}{7} + \frac{x^{3}}{3} + C}\)
1(iv) \(\displaystyle \int \dfrac{12x^{2}-7x-10}{3x+2}\,dx\)
\((3x+2)(4x-5)=12x^{2}-7x-10\Rightarrow\dfrac{12x^{2}-7x-10}{3x+2}=4x-5\)
\(\Rightarrow \int(4x-5)\,dx=2x^{2}-5x+C\)
\(\boxed{2x^{2}-5x+C}\)
1(v) \(\displaystyle \int \dfrac{x^{3}-6x-9}{x-3}\,dx\)
Divide: \(x^{3}-6x-9=(x-3)(x^{2}+3x+3)\)
\(\Rightarrow \dfrac{x^{3}-6x-9}{x-3}=x^{2}+3x+3\)
\(\Rightarrow \int(x^{2}+3x+3)\,dx=\frac{x^{3}}{3}+\frac{3x^{2}}{2}+3x+C\)
\(\boxed{\frac{x^{3}}{3}+\frac{3x^{2}}{2}+3x+C}\)
1(vi) \(\displaystyle \int \dfrac{x^{6}-1}{x-1}\,dx\)
\(\dfrac{x^{6}-1}{x-1}=x^{5}+x^{4}+x^{3}+x^{2}+x+1\)
\(\Rightarrow \int(x^{5}+x^{4}+x^{3}+x^{2}+x+1)\,dx=\frac{x^{6}}{6}+\frac{x^{5}}{5}+\frac{x^{4}}{4}+\frac{x^{3}}{3}+\frac{x^{2}}{2}+x+C\)
\(\boxed{\frac{x^{6}}{6}+\frac{x^{5}}{5}+\frac{x^{4}}{4}+\frac{x^{3}}{3}+\frac{x^{2}}{2}+x+C}\)
1(vii) \(\displaystyle \int \dfrac{x^{3}-4x^{2}+5x-2}{(x-1)^{2}}\,dx\)
First, we have to factorise the numerator:
\((x^{3}-4x^{2}+5x-2)\)
\(= x^{3}-x^{2}-3x^{2}+3x+2x-2\)
\(= x^{2}(x-1) – 3x(x-1) + 2(x-1)\)
\(= (x-1)(x^{2}-3x+2)\)
\(= (x-1)\{x^{2}-2x – x + 2\}\)
\(= (x-1)\{x(x-2) – 1(x-2)\}\)
\(= (x-1)(x-2)(x-1)\)
\(= (x-1)^{2}(x-2)\)
∴ \(\displaystyle \int \dfrac{x^{3}-4x^{2}+5x-2}{(x-1)^{2}}\,dx\)
\(= \int \dfrac{(x-1)^{2}(x-2)}{(x-1)^{2}}\,dx\)
\(= \int (x-2)\,dx\)
\(= \int x\,dx – 2\int dx\)
\(= \dfrac{x^{2}}{2} – 2x + C\)
\(\boxed{\dfrac{x^{2}}{2} – 2x + C}\)
1(viii) \(\displaystyle \int \dfrac{x^{4}+x^{2}+1}{x^{2}-x+1}\,dx\)
\(x^{4}+x^{2}+1=(x^{2}-x+1)(x^{2}+x+1)\)
\(\Rightarrow \dfrac{x^{4}+x^{2}+1}{x^{2}-x+1}=x^{2}+x+1\)
\(\Rightarrow \int(x^{2}+x+1)\,dx=\frac{x^{3}}{3}+\frac{x^{2}}{2}+x+C\)
\(\boxed{\frac{x^{3}}{3}+\frac{x^{2}}{2}+x+C}\)
1(ix) \(\displaystyle \int \dfrac{2x^{3}-7x+2}{x^{3}+2x^{2}}\,dx\)
Denominator \(=x^{2}(x+2)\)
\(x=-2\Rightarrow 0\Rightarrow (x+2)\) factor of numerator
Divide \(2x^{3}-7x+2=(x+2)(2x^{2}-4x+1)\)
\(\Rightarrow \dfrac{2x^{3}-7x+2}{x^{2}(x+2)}=\dfrac{2x^{2}-4x+1}{x^{2}}=2-\dfrac{4}{x}+\dfrac{1}{x^{2}}\)
\(\Rightarrow \int(2-\dfrac{4}{x}+\dfrac{1}{x^{2}})\,dx=2x-4\ln|x|-\dfrac{1}{x}+C\)
\(\boxed{2x-4\ln|x|-\dfrac{1}{x}+C}\)
2(i) \(\displaystyle \int \dfrac{e^{6\log x}-e^{4\log x}}{e^{3\log x}-e^{\log x}}\,dx\)
\(e^{k\log x}=x^{k}\Rightarrow\dfrac{x^{6}-x^{4}}{x^{3}-x}=\dfrac{x^{4}(x^{2}-1)}{x(x^{2}-1)}=x^{3}\)
\(\Rightarrow \int x^{3}\,dx=\frac{x^{4}}{4}+C\)
\(\boxed{\frac{x^{4}}{4}+C}\)
2(ii) \(\displaystyle \int \frac{e^{2x-1}-e^{1-2x}}{e^{x+2}}\,dx\)
\(\dfrac{e^{2x-1}-e^{1-2x}}{e^{x+2}}
= e^{(2x-1)-(x+2)} – e^{(1-2x)-(x+2)}\)
= e^{(2x-1)-(x+2)} – e^{(1-2x)-(x+2)}\)
\(\ = e^{x-3} – e^{-3x-1}\)
\(\displaystyle \int\big(e^{x-3}-e^{-3x-1}\big)\,dx
= \int e^{x-3}\,dx – \int e^{-3x-1}\,dx\)
= \int e^{x-3}\,dx – \int e^{-3x-1}\,dx\)
\(\displaystyle \int e^{x-3}\,dx = e^{-3}\int e^{x}\,dx = e^{x-3}\)
\(\displaystyle \int e^{-3x-1}\,dx = e^{-1}\int e^{-3x}\,dx
= e^{-1}\cdot\frac{e^{-3x}}{-3} = -\tfrac{1}{3}e^{-3x-1}\)
= e^{-1}\cdot\frac{e^{-3x}}{-3} = -\tfrac{1}{3}e^{-3x-1}\)
\(\Rightarrow \displaystyle \int \frac{e^{2x-1}-e^{1-2x}}{e^{x+2}}\,dx
= e^{x-3} + \tfrac{1}{3}e^{-3x-1} + C\)
= e^{x-3} + \tfrac{1}{3}e^{-3x-1} + C\)
\(\boxed{\,e^{x-3} + \tfrac{1}{3}e^{-3x-1} + C\,}\)
3(ii) \(\displaystyle \int \dfrac{27^{1+x}+9^{1-x}}{3^{x}}\,dx\)
\(27^{1+x}=27\cdot3^{3x},\ 9^{1-x}=9\cdot3^{-2x}\)
\(\Rightarrow \dfrac{27\cdot3^{3x}+9\cdot3^{-2x}}{3^{x}}=27\cdot3^{2x}+9\cdot3^{-3x}\)
\(\Rightarrow \int(27\cdot3^{2x}+9\cdot3^{-3x})\,dx
=\frac{27\cdot3^{2x}}{2\ln3}-\frac{3\cdot3^{-3x}}{\ln3}+C\)
=\frac{27\cdot3^{2x}}{2\ln3}-\frac{3\cdot3^{-3x}}{\ln3}+C\)
\(\boxed{\frac{27\cdot3^{2x}}{2\ln3}-\frac{3\cdot3^{-3x}}{\ln3}+C}\)
3(iii) \(\displaystyle \int \dfrac{2^{6x}-1}{2^{2x}-1}\,dx\)
Let \(t=2^{2x}\Rightarrow2^{6x}=t^{3}\)
\(\Rightarrow\dfrac{t^{3}-1}{t-1}=t^{2}+t+1\)
\(dt=2\ln2\,t\,dx\Rightarrow dx=\dfrac{dt}{2\ln2\,t}\)
\(\Rightarrow \int(t^{2}+t+1)\,dx=\frac{1}{2\ln2}\int(t+1+\tfrac{1}{t})\,dt\)
\(\Rightarrow \frac{1}{2\ln2}(\frac{t^{2}}{2}+t+\ln|t|)+C\)
Substitute \(t=2^{2x}\Rightarrow\boxed{\frac{2^{4x}}{4\ln2}+\frac{2^{2x}}{2\ln2}+x+C}\)
\(\boxed{\frac{2^{4x}}{4\ln2}+\frac{2^{2x}}{2\ln2}+x+C}\)
1. \( \displaystyle \int \sin x\,\sin 2x\,\cos 3x\,dx \)
Use \( \sin A\sin B=\tfrac12[\cos(A-B)-\cos(A+B)]\).
\[
\sin x\sin2x=\tfrac12(\cos x-\cos3x).
\]
\[
\text{Integrand } = \tfrac12(\cos x-\cos3x)\cos3x
= \tfrac12(\cos x\cos3x – \cos^2 3x).
\]
\[
\cos x\cos3x=\tfrac12(\cos2x+\cos4x),\qquad \cos^2 3x=\tfrac{1+\cos6x}{2}.
\]
\[
\text{Integrand }= \tfrac14(\cos2x+\cos4x-1-\cos6x).
\]
\[
\int = \tfrac14\Big(\tfrac{\sin2x}{2}+\tfrac{\sin4x}{4}-x-\tfrac{\sin6x}{6}\Big)+C.
\]
\( \boxed{\dfrac{1}{4}\Big(\dfrac{\sin2x}{2}+\dfrac{\sin4x}{4}-\dfrac{\sin6x}{6}-x\Big)+C} \)
2. \( \displaystyle \int \cos x\,\cos2x\,\cos3x\,dx \)
\( \cos x\cos2x=\tfrac12(\cos x+\cos3x) \Rightarrow \) integrand \(=\tfrac12(\cos x\cos3x + \cos^2 3x).\)
\[
\cos x\cos3x=\tfrac12(\cos2x+\cos4x),\qquad \cos^2 3x=\tfrac{1+\cos6x}{2}.
\]
\[
\text{Integrand }= \tfrac14(\cos2x+\cos4x+1+\cos6x).
\]
\[
\int = \tfrac14\Big(x+\tfrac{\sin2x}{2}+\tfrac{\sin4x}{4}+\tfrac{\sin6x}{6}\Big)+C.
\]
\( \boxed{\dfrac{1}{4}\Big(x+\dfrac{\sin2x}{2}+\dfrac{\sin4x}{4}+\dfrac{\sin6x}{6}\Big)+C} \)
3. \( \displaystyle \int \sin x\,\cos2x\,\cos3x\,dx \)
\( \cos2x\cos3x=\tfrac12(\cos x+\cos5x) \Rightarrow \) integrand \(=\tfrac12(\sin x\cos x+\sin x\cos5x).\)
\[
\sin x\cos x=\tfrac12\sin2x,\quad \sin x\cos5x=\tfrac12(\sin6x-\sin4x).
\]
\[
\text{Integrand }=\tfrac14(\sin2x+\sin6x-\sin4x).
\]
\[
\int = \tfrac14\Big(-\tfrac{\cos2x}{2}-\tfrac{\cos6x}{6}+\tfrac{\cos4x}{4}\Big)+C.
\]
\[
\sin x\cos x=\tfrac12\sin2x,\quad \sin x\cos5x=\tfrac12(\sin6x-\sin4x).
\]
\[
\text{Integrand }=\tfrac14(\sin2x+\sin6x-\sin4x).
\]
\[
\int = \tfrac14\Big(-\tfrac{\cos2x}{2}-\tfrac{\cos6x}{6}+\tfrac{\cos4x}{4}\Big)+C.
\]
\( \boxed{-\dfrac{1}{4}\Big(\dfrac{\cos2x}{2}-\dfrac{\cos4x}{4}+\dfrac{\cos6x}{6}\Big)+C} \)
4. \( \displaystyle \int \sin x\,\sin2x\,\sin3x\,dx \)
\( \sin2x\sin3x=\tfrac12(\cos x-\cos5x) \Rightarrow \) integrand \(=\tfrac12(\sin x\cos x-\sin x\cos5x).\)
\[
\sin x\cos x=\tfrac12\sin2x,\quad \sin x\cos5x=\tfrac12(\sin6x-\sin4x).
\]
\[
\text{Integrand }=\tfrac14(\sin2x+\sin4x-\sin6x).
\]
\[
\int=\tfrac14\Big(-\tfrac{\cos2x}{2}-\tfrac{\cos4x}{4}+\tfrac{\cos6x}{6}\Big)+C
= -\tfrac{1}{48}(6\cos2x+3\cos4x-2\cos6x)+C.
\]
\[
\sin x\cos x=\tfrac12\sin2x,\quad \sin x\cos5x=\tfrac12(\sin6x-\sin4x).
\]
\[
\text{Integrand }=\tfrac14(\sin2x+\sin4x-\sin6x).
\]
\[
\int=\tfrac14\Big(-\tfrac{\cos2x}{2}-\tfrac{\cos4x}{4}+\tfrac{\cos6x}{6}\Big)+C
= -\tfrac{1}{48}(6\cos2x+3\cos4x-2\cos6x)+C.
\]
\( \boxed{-\dfrac{1}{48}(6\cos2x+3\cos4x-2\cos6x)+C} \)
5. \( \displaystyle \int \sin^{2}x\,\cos2x\,dx \)
\( \sin^{2}x=\tfrac{1-\cos2x}{2} \Rightarrow \sin^{2}x\cos2x=\tfrac12\cos2x – \tfrac12\cos^22x.\)
\[
\cos^22x=\tfrac{1+\cos4x}{2}\Rightarrow \text{integrand }=\tfrac12\cos2x – \tfrac14 – \tfrac14\cos4x.
\]
\[
\int = \tfrac{\sin2x}{4} – \tfrac{x}{4} – \tfrac{\sin4x}{16} + C
= \tfrac14\Big(\sin2x – \tfrac{\sin4x}{4} – x\Big)+C.
\]
\[
\cos^22x=\tfrac{1+\cos4x}{2}\Rightarrow \text{integrand }=\tfrac12\cos2x – \tfrac14 – \tfrac14\cos4x.
\]
\[
\int = \tfrac{\sin2x}{4} – \tfrac{x}{4} – \tfrac{\sin4x}{16} + C
= \tfrac14\Big(\sin2x – \tfrac{\sin4x}{4} – x\Big)+C.
\]
\( \boxed{\dfrac{1}{4}\Big(\sin2x – \dfrac{\sin4x}{4} – x\Big) + C} \)
6. \( \displaystyle \int 2\cos^{2}2x\,\sin4x\,dx \)
Use \( \sin4x=2\sin2x\cos2x.\) Then integrand \(= 2\cos^{2}2x \cdot 2\sin2x\cos2x = 4\cos^{3}2x\sin2x.\)
Put \( t=\cos2x\Rightarrow dt=-2\sin2x\,dx.\)
\[
I = 4\int t^{3}\sin2x\,dx = 4\int t^{3}\Big(-\tfrac12dt\Big) = -2\int t^{3}dt = -\tfrac12 t^{4}+C.
\]
\[
\therefore I = -\tfrac12\cos^{4}2x + C.
\]
Put \( t=\cos2x\Rightarrow dt=-2\sin2x\,dx.\)
\[
I = 4\int t^{3}\sin2x\,dx = 4\int t^{3}\Big(-\tfrac12dt\Big) = -2\int t^{3}dt = -\tfrac12 t^{4}+C.
\]
\[
\therefore I = -\tfrac12\cos^{4}2x + C.
\]
\( \boxed{-\dfrac{1}{2}\cos^{4}2x + C} \)
7. \( \displaystyle \int \frac{\cos2x – \cos2\alpha}{\cos x – \cos\alpha}\,dx \)
Use \( \cos2\theta=2\cos^{2}\theta-1.\)
\[
\cos2x-\cos2\alpha = 2(\cos^{2}x-\cos^{2}\alpha) = 2(\cos x-\cos\alpha)(\cos x+\cos\alpha).
\]
\[
\therefore \frac{\cos2x-\cos2\alpha}{\cos x-\cos\alpha}=2(\cos x+\cos\alpha).
\]
\[
I= \int 2(\cos x+\cos\alpha)\,dx = 2\sin x + 2x\cos\alpha + C.
\]
\[
\cos2x-\cos2\alpha = 2(\cos^{2}x-\cos^{2}\alpha) = 2(\cos x-\cos\alpha)(\cos x+\cos\alpha).
\]
\[
\therefore \frac{\cos2x-\cos2\alpha}{\cos x-\cos\alpha}=2(\cos x+\cos\alpha).
\]
\[
I= \int 2(\cos x+\cos\alpha)\,dx = 2\sin x + 2x\cos\alpha + C.
\]
\( \boxed{2(\sin x + x\cos\alpha)+C} \)
8. \( \displaystyle \int \frac{\sin x – \cos2x}{1+\sin x}\,dx \)
Use \( \cos2x=1-2\sin^{2}x.\)
\[
\sin x – \cos2x = \sin x – (1-2\sin^{2}x) = 2\sin^{2}x + \sin x – 1.
\]
Factor (put \( s=\sin x \)):
\[
2s^{2}+s-1=(2s-1)(s+1).
\]
\[
\text{So } \frac{\sin x – \cos2x}{1+\sin x} = \frac{(2s-1)(s+1)}{s+1} = 2s-1 = 2\sin x -1.
\]
\[
I=\int(2\sin x – 1)\,dx = -2\cos x – x + C.
\]
\[
\sin x – \cos2x = \sin x – (1-2\sin^{2}x) = 2\sin^{2}x + \sin x – 1.
\]
Factor (put \( s=\sin x \)):
\[
2s^{2}+s-1=(2s-1)(s+1).
\]
\[
\text{So } \frac{\sin x – \cos2x}{1+\sin x} = \frac{(2s-1)(s+1)}{s+1} = 2s-1 = 2\sin x -1.
\]
\[
I=\int(2\sin x – 1)\,dx = -2\cos x – x + C.
\]
\( \boxed{-(x+2\cos x)+C} \)
9. \( \displaystyle \int \frac{\cos x – \cos2x}{1 – \cos x}\,dx \)
Use \( \cos2x=2\cos^{2}x-1.\)
\[
\cos x – \cos2x = \cos x – (2\cos^{2}x -1) = 1 + \cos x – 2\cos^{2}x.
\]
Put \( c=\cos x\): \(1 + c – 2c^{2}=(1-c)(1+2c).\)
\[
\therefore \frac{\cos x – \cos2x}{1-\cos x} = \frac{(1-c)(1+2c)}{1-c}=1+2c=1+2\cos x.
\]
\[
I=\int(1+2\cos x)\,dx = x + 2\sin x + C.
\]
\[
\cos x – \cos2x = \cos x – (2\cos^{2}x -1) = 1 + \cos x – 2\cos^{2}x.
\]
Put \( c=\cos x\): \(1 + c – 2c^{2}=(1-c)(1+2c).\)
\[
\therefore \frac{\cos x – \cos2x}{1-\cos x} = \frac{(1-c)(1+2c)}{1-c}=1+2c=1+2\cos x.
\]
\[
I=\int(1+2\cos x)\,dx = x + 2\sin x + C.
\]
\( \boxed{x+2\sin x + C} \)
10. \( \displaystyle \int \frac{\sin^{8}x – \cos^{8}x}{1 – 2\sin^{2}x\cos^{2}x}\,dx \)
Factor:
\[
\sin^{8}x – \cos^{8}x = (\sin^{4}x – \cos^{4}x)(\sin^{4}x + \cos^{4}x).
\]
Now \( \sin^{4}-\cos^{4}=(\sin^{2}-\cos^{2})(\sin^{2}+\cos^{2})=\sin^{2}-\cos^{2}.\)
Also \( \sin^{4}+\cos^{4} = 1 – 2\sin^{2}x\cos^{2}x.\)
\[
\therefore \frac{\sin^{8}x – \cos^{8}x}{1-2\sin^{2}x\cos^{2}x} = \sin^{2}x – \cos^{2}x = -\cos2x.
\]
\[
I = \int -\cos2x\,dx = -\tfrac12\sin2x + C.
\]
\[
\sin^{8}x – \cos^{8}x = (\sin^{4}x – \cos^{4}x)(\sin^{4}x + \cos^{4}x).
\]
Now \( \sin^{4}-\cos^{4}=(\sin^{2}-\cos^{2})(\sin^{2}+\cos^{2})=\sin^{2}-\cos^{2}.\)
Also \( \sin^{4}+\cos^{4} = 1 – 2\sin^{2}x\cos^{2}x.\)
\[
\therefore \frac{\sin^{8}x – \cos^{8}x}{1-2\sin^{2}x\cos^{2}x} = \sin^{2}x – \cos^{2}x = -\cos2x.
\]
\[
I = \int -\cos2x\,dx = -\tfrac12\sin2x + C.
\]
\( \boxed{-\dfrac{1}{2}\sin2x + C} \)
11. \( \displaystyle \int \frac{\cos5x + \cos4x}{1 – 2\cos3x}\,dx \)
\[
\begin{aligned}
I &= \int \frac{\cos5x+\cos4x}{1-2\cos3x}\,dx \\[6pt]
&= \int \frac{\sin3x\big(\cos5x+\cos4x\big)}{\sin3x\big(1-2\cos3x\big)}\,dx \\[6pt]
\cos5x+\cos4x &= 2\cos\!\left(\tfrac{9x}{2}\right)\cos\!\left(\tfrac{x}{2}\right)\\[6pt]
\Rightarrow I &= \int \frac{\sin3x\cdot 2\cos\!\left(\tfrac{9x}{2}\right)\cos\!\left(\tfrac{x}{2}\right)}
{\sin3x – 2\sin3x\cos3x}\,dx \\[8pt]
\sin3x &= 2\sin\!\left(\tfrac{3x}{2}\right)\cos\!\left(\tfrac{3x}{2}\right),\qquad
\sin6x = 2\sin3x\cos3x\\[6pt]
\sin3x – \sin6x &= \sin3x – 2\sin3x\cos3x = \sin3x(1-2\cos3x)\\[6pt]
\sin3x-\sin6x &= -2\cos\!\left(\tfrac{9x}{2}\right)\sin\!\left(\tfrac{3x}{2}\right)
\quad\text{(use } \sin A-\sin B = 2\cos\!\tfrac{A+B}{2}\sin\!\tfrac{A-B}{2})\\[8pt]
\Rightarrow I &= \int
\frac{2\cdot 2\sin\!\left(\tfrac{3x}{2}\right)\cos\!\left(\tfrac{3x}{2}\right)
\cos\!\left(\tfrac{9x}{2}\right)\cos\!\left(\tfrac{x}{2}\right)}
{-2\cos\!\left(\tfrac{9x}{2}\right)\sin\!\left(\tfrac{3x}{2}\right)}\,dx \\[8pt]
&= \int -2\cos\!\left(\tfrac{3x}{2}\right)\cos\!\left(\tfrac{x}{2}\right)\,dx \\[8pt]
-2\cos A\cos B &= -\big(\cos(A+B)+\cos(A-B)\big)\\[6pt]
\Rightarrow I &= -\int\big(\cos2x+\cos x\big)\,dx \\[6pt]
&= -\Big(\tfrac{\sin2x}{2} + \sin x\Big) + C
\end{aligned}
\]
\begin{aligned}
I &= \int \frac{\cos5x+\cos4x}{1-2\cos3x}\,dx \\[6pt]
&= \int \frac{\sin3x\big(\cos5x+\cos4x\big)}{\sin3x\big(1-2\cos3x\big)}\,dx \\[6pt]
\cos5x+\cos4x &= 2\cos\!\left(\tfrac{9x}{2}\right)\cos\!\left(\tfrac{x}{2}\right)\\[6pt]
\Rightarrow I &= \int \frac{\sin3x\cdot 2\cos\!\left(\tfrac{9x}{2}\right)\cos\!\left(\tfrac{x}{2}\right)}
{\sin3x – 2\sin3x\cos3x}\,dx \\[8pt]
\sin3x &= 2\sin\!\left(\tfrac{3x}{2}\right)\cos\!\left(\tfrac{3x}{2}\right),\qquad
\sin6x = 2\sin3x\cos3x\\[6pt]
\sin3x – \sin6x &= \sin3x – 2\sin3x\cos3x = \sin3x(1-2\cos3x)\\[6pt]
\sin3x-\sin6x &= -2\cos\!\left(\tfrac{9x}{2}\right)\sin\!\left(\tfrac{3x}{2}\right)
\quad\text{(use } \sin A-\sin B = 2\cos\!\tfrac{A+B}{2}\sin\!\tfrac{A-B}{2})\\[8pt]
\Rightarrow I &= \int
\frac{2\cdot 2\sin\!\left(\tfrac{3x}{2}\right)\cos\!\left(\tfrac{3x}{2}\right)
\cos\!\left(\tfrac{9x}{2}\right)\cos\!\left(\tfrac{x}{2}\right)}
{-2\cos\!\left(\tfrac{9x}{2}\right)\sin\!\left(\tfrac{3x}{2}\right)}\,dx \\[8pt]
&= \int -2\cos\!\left(\tfrac{3x}{2}\right)\cos\!\left(\tfrac{x}{2}\right)\,dx \\[8pt]
-2\cos A\cos B &= -\big(\cos(A+B)+\cos(A-B)\big)\\[6pt]
\Rightarrow I &= -\int\big(\cos2x+\cos x\big)\,dx \\[6pt]
&= -\Big(\tfrac{\sin2x}{2} + \sin x\Big) + C
\end{aligned}
\]
\[
\boxed{\;-\,\Big(\tfrac{\sin2x}{2}+\sin x\Big) + C\;}
\]
\boxed{\;-\,\Big(\tfrac{\sin2x}{2}+\sin x\Big) + C\;}
\]
12. \( \displaystyle \int \sec^{2}x\,\cos^{2}2x\,dx \)
\[
\begin{aligned}
I &= \int \sec^{2}x\,\cos^{2}2x\,dx \\[6pt]
&= \int \frac{\cos^{2}2x}{\cos^{2}x}\,dx \\[8pt]
\text{Use } \cos2x &= 2\cos^{2}x – 1 \\[6pt]
\cos^{2}2x &= (2\cos^{2}x-1)^{2} = 4\cos^{4}x – 4\cos^{2}x + 1 \\[8pt]
\therefore \frac{\cos^{2}2x}{\cos^{2}x}
&= \frac{4\cos^{4}x – 4\cos^{2}x + 1}{\cos^{2}x}
= 4\cos^{2}x – 4 + \frac{1}{\cos^{2}x} \\[8pt]
I &= \int \big(4\cos^{2}x – 4 + \sec^{2}x\big)\,dx \\[8pt]
&= 4\int \cos^{2}x\,dx – 4\int dx + \int \sec^{2}x\,dx \\[6pt]
\cos^{2}x &= \tfrac{1+\cos2x}{2} \;\Rightarrow\; \int \cos^{2}x\,dx = \tfrac{x}{2}+\tfrac{\sin2x}{4} \\[8pt]
\Rightarrow I &= 4\Big(\tfrac{x}{2}+\tfrac{\sin2x}{4}\Big) – 4x + \tan x + C \\[8pt]
&= 2x + \sin2x – 4x + \tan x + C \\[6pt]
&= -2x + \sin2x + \tan x + C
\end{aligned}
\]
\begin{aligned}
I &= \int \sec^{2}x\,\cos^{2}2x\,dx \\[6pt]
&= \int \frac{\cos^{2}2x}{\cos^{2}x}\,dx \\[8pt]
\text{Use } \cos2x &= 2\cos^{2}x – 1 \\[6pt]
\cos^{2}2x &= (2\cos^{2}x-1)^{2} = 4\cos^{4}x – 4\cos^{2}x + 1 \\[8pt]
\therefore \frac{\cos^{2}2x}{\cos^{2}x}
&= \frac{4\cos^{4}x – 4\cos^{2}x + 1}{\cos^{2}x}
= 4\cos^{2}x – 4 + \frac{1}{\cos^{2}x} \\[8pt]
I &= \int \big(4\cos^{2}x – 4 + \sec^{2}x\big)\,dx \\[8pt]
&= 4\int \cos^{2}x\,dx – 4\int dx + \int \sec^{2}x\,dx \\[6pt]
\cos^{2}x &= \tfrac{1+\cos2x}{2} \;\Rightarrow\; \int \cos^{2}x\,dx = \tfrac{x}{2}+\tfrac{\sin2x}{4} \\[8pt]
\Rightarrow I &= 4\Big(\tfrac{x}{2}+\tfrac{\sin2x}{4}\Big) – 4x + \tan x + C \\[8pt]
&= 2x + \sin2x – 4x + \tan x + C \\[6pt]
&= -2x + \sin2x + \tan x + C
\end{aligned}
\]
\[
\boxed{\; -2x + \sin2x + \tan x + C \;}
\]
\boxed{\; -2x + \sin2x + \tan x + C \;}
\]
13. \( \displaystyle \int \frac{\cos^{4}x}{\sin^{2}x}\,dx \)
Write \( \cos^{4}x=(1-\sin^{2}x)^{2}=1-2\sin^{2}x+\sin^{4}x.\)
\[
\frac{\cos^{4}x}{\sin^{2}x} = \frac{1}{\sin^{2}x} – 2 + \sin^{2}x = \csc^{2}x – 2 + \sin^{2}x.
\]
Integrate termwise:
\[
\int \csc^{2}x\,dx = -\cot x,\quad \int -2\,dx = -2x,\quad \int \sin^{2}x\,dx = \tfrac{x}{2} – \tfrac{\sin2x}{4}.
\]
\[
I = -\cot x – 2x + \tfrac{x}{2} – \tfrac{\sin2x}{4} + C = -\cot x – \tfrac{3x}{2} – \tfrac{\sin2x}{4} + C.
\]
\[
\frac{\cos^{4}x}{\sin^{2}x} = \frac{1}{\sin^{2}x} – 2 + \sin^{2}x = \csc^{2}x – 2 + \sin^{2}x.
\]
Integrate termwise:
\[
\int \csc^{2}x\,dx = -\cot x,\quad \int -2\,dx = -2x,\quad \int \sin^{2}x\,dx = \tfrac{x}{2} – \tfrac{\sin2x}{4}.
\]
\[
I = -\cot x – 2x + \tfrac{x}{2} – \tfrac{\sin2x}{4} + C = -\cot x – \tfrac{3x}{2} – \tfrac{\sin2x}{4} + C.
\]
\( \boxed{-\cot x – \dfrac{3x}{2} – \dfrac{\sin2x}{4} + C} \)
14. \( \displaystyle \int \sin(mx)\sin(nx)\,dx \quad (m^{2}\ne n^{2}) \)
Use identity \( \sin A\sin B=\tfrac12[\cos(A-B)-\cos(A+B)].\)
\[
\sin(mx)\sin(nx)=\tfrac12[\cos((m-n)x)-\cos((m+n)x)].
\]
Integrate:
\[
I=\tfrac12\Big(\frac{\sin((m-n)x)}{m-n}-\frac{\sin((m+n)x)}{m+n}\Big)+C.
\]
\[
\sin(mx)\sin(nx)=\tfrac12[\cos((m-n)x)-\cos((m+n)x)].
\]
Integrate:
\[
I=\tfrac12\Big(\frac{\sin((m-n)x)}{m-n}-\frac{\sin((m+n)x)}{m+n}\Big)+C.
\]
\( \boxed{\tfrac12\Big(\dfrac{\sin((m-n)x)}{m-n}-\dfrac{\sin((m+n)x)}{m+n}\Big)+C} \)
15 i. Find \( f(x) \) such that \( f'(x)=\dfrac{x^{4}-1}{x(x-1)} \)
Divide polynomial:
\[
\frac{x^{4}-1}{x(x-1)} = \frac{x^{4}-1}{x^{2}-x} = x^{2}+x+1 + \frac{1}{x}.
\]
(Check: multiply back \(x(x-1)(x^{2}+x+1+\tfrac1x) = x^{4}-1\).)
Integrate:
\[
f(x)=\int\Big(x^{2}+x+1+\frac{1}{x}\Big)dx = \frac{x^{3}}{3}+\frac{x^{2}}{2}+x+\ln x + C.
\]
\[
\frac{x^{4}-1}{x(x-1)} = \frac{x^{4}-1}{x^{2}-x} = x^{2}+x+1 + \frac{1}{x}.
\]
(Check: multiply back \(x(x-1)(x^{2}+x+1+\tfrac1x) = x^{4}-1\).)
Integrate:
\[
f(x)=\int\Big(x^{2}+x+1+\frac{1}{x}\Big)dx = \frac{x^{3}}{3}+\frac{x^{2}}{2}+x+\ln x + C.
\]
\( \boxed{\dfrac{x^{3}}{3}+\dfrac{x^{2}}{2}+x+\ln x + C} \)
15(ii). Find \( f(x) \) such that \( f'(x) = 2^{2x-1} + 2^{1-2x} \).
\[
\begin{aligned}
f(x) &= \int \big(2^{2x-1} + 2^{1-2x}\big)\,dx \\[8pt]
&= \tfrac{1}{2}\int 2^{2x}\,dx + 2\int 2^{-2x}\,dx \\[8pt]
&= \tfrac{1}{2}\cdot\frac{2^{2x}}{2\ln2} + 2\cdot\frac{2^{-2x}}{-2\ln2} + C \\[8pt]
&= \frac{2^{2x}}{4\ln2} – \frac{2^{-2x}}{\ln2} + C \\[8pt]
&= \frac{1}{4\ln2}\big(2^{2x} – 4\cdot2^{-2x}\big) + C
\end{aligned}
\]
\begin{aligned}
f(x) &= \int \big(2^{2x-1} + 2^{1-2x}\big)\,dx \\[8pt]
&= \tfrac{1}{2}\int 2^{2x}\,dx + 2\int 2^{-2x}\,dx \\[8pt]
&= \tfrac{1}{2}\cdot\frac{2^{2x}}{2\ln2} + 2\cdot\frac{2^{-2x}}{-2\ln2} + C \\[8pt]
&= \frac{2^{2x}}{4\ln2} – \frac{2^{-2x}}{\ln2} + C \\[8pt]
&= \frac{1}{4\ln2}\big(2^{2x} – 4\cdot2^{-2x}\big) + C
\end{aligned}
\]
\[
\boxed{\, f(x) = \dfrac{2^{2x}}{4\ln2} – \dfrac{2^{-2x}}{\ln2} + C \,}
\]
\boxed{\, f(x) = \dfrac{2^{2x}}{4\ln2} – \dfrac{2^{-2x}}{\ln2} + C \,}
\]