\(I=\int _{ e }^{ { e }^{ 2 } }{ \frac { dx }{ xlogx } } \)
Put log x=t
\(\Rightarrow \frac { dx }{ t } =dt\)
On Integration
log x=t
\(loge=t\Rightarrow 1=t\)
\(\Rightarrow \) 1=t lower limit
log e²=t
\(\Rightarrow \) 2=t upper limit
\(I=\int _{ 1 }^{ 2 }{ \frac { dx }{ t } } \)
\(\Rightarrow I=\left[ logt \right] ^{ 2 }_{ 1 }\)
\(\Rightarrow I=\left[ log2-log1 \right] \)
\(\Rightarrow I=\left[ log2-0 \right] \)
\(\Rightarrow I=log2\)

\(I=\int _{ a }^{ b }{ \frac { logx }{ x } } dx\)
Put log x=t
\(\frac { dx }{ x } =dt\)
log b=t upper limit
log a=t lower limit
\(=\int _{ loga }^{ logb }{ tdt } \)
\(\Rightarrow I=\left[ \frac { { t }^{ 2 } }{ 2 } \right] ^{ logb }_{ loga }\)
\(\Rightarrow I=\frac { 1 }{ 2 } \left[ (logb)^{ 2 }-(loga)^{ 2 } \right] \)
\(\Rightarrow I=\frac { 1 }{ 2 } \left[ (logb+loga)(logb-loga) \right] \)
\(\Rightarrow I=\frac { 1 }{ 2 } \left[ (log\quad ab)\left( log\frac { b }{ a } \right) \right] \)

\(\int { { e }^{ x }cotx } dx+\int { { e }^{ x }(logsinx) } dx\)
\(=\int { { e }^{ x } } cotxdx+log(sinx)\int { { e }^{ x }dx } -\int { \left[ \frac { d }{ dx } (logsinx)\int { { e }^{ x }dx } \right] } \)
\(=\int { { e }^{ x }cotx } dx-{ e }^{ x }log(sinx)-\int { cotx{ e }^{ x }dx } \)
\(={ e }^{ x }log\left| sinx \right| +C\)

\(\int { \left[ \frac { 2sinxcosx }{ 2 } \right] ^{ 2 } } dx\)
\(=\frac { 1 }{ 4 } \int { (sin2x)^{ 2 } } dx\)
\(=\frac { 1 }{ 4 } \int { { sin }^{ 2 }2x } dx\)
\(\frac { 1 }{ 4 } \int { \frac { 1-cos4x }{ 2 } } dx=\frac { 1 }{ 8 } \int { 1dx } -\frac { 1 }{ 8 } \int { cos4xdx } \)
 \(=\frac { x }{ 8 } -\frac { 1 }{ 8 } \frac { sin4x }{ 4 } \)
\(=\frac { 1 }{ 8 } \left( x-\frac { 1 }{ 4 } sin4x \right) +c\)

\(\int { { cos }^{ 3 } } xdx=\frac { 1 }{ 4 } \int { (cos } 3x+3cosx)dx\)
\([\because cos3x=4cos^{ 3 }x-3cosx]\)
\(=\frac { 1 }{ 4 } \int { cos3x } dx+\int { \frac { 3 }{ 4 } } cosxdx\)
\(=\frac { 1 }{ 4 } \left( \frac { sin3x }{ 3 } \right) +\frac { 3 }{ 4 } sinx+C\)
 \(=\frac { 1 }{ 12 } sin3x+\frac { 3 }{ 4 } sinx+C\)

\(\int \sqrt{\tan x} \cdot \sec ^{2} x d x=\int \sqrt{t} d t=\frac{2}{3} t^{\frac{3}{2}}+C=\frac{2}{3}(\tan x)^{\frac{3}{2}}+C \)

\(\text { We have }\int \tan ^{-1}(\cot x) d x=\int \tan ^{-1}\left\{\tan \left(\frac{\pi}{2}-x\right)\right\} d x=\int\left(\frac{\pi}{2}-x\right) d x=\frac{\pi}{2} x-\frac{x^{2}}{2}+C \)

\(f(x)=\frac{(a x+b)^{3}}{3 a} \text { as } \int(a x+b)^{2} d x=\frac{(a x+b)^{3}}{3 a}+C\)

\(={1\over4}tan^{-1}({x\over4})\ +\ c\)

\(\int \frac{x^{2}+4 x}{x^{3}+6 x^{2}+5} d x =\frac{1}{3} \int \frac{1}{t} d t \)
\(=\frac{1}{3} \log |t|+C=\frac{1}{3} \log \left|x^{3}+6 x^{2}+5\right|+C\)

Let \( \int { \frac { { x }^{ 2 }+x+1 }{ ({ x }^{ 2 }+1)(x+2) } } dx.\)
Let \(I= \int { \frac { { x }^{ 2 }+x+1 }{ ({ x }^{ 2 }+1)(x+2) } } =\frac { Ax+B }{ { x }^{ 2 }+1 } +\frac { C }{ x+2 } ....(1)\)
Multiplying by \(({ x }^{ 2 }+1)(x+2)\) 
we get \({ x }^{ 2 }+x+1=(Ax+B)(x+2)+C({ x }^{ 2 }+1)\)
Putting \(x=-2,4-2+1=0+C(4+1)\)
\(\Rightarrow C=\frac { 3 }{ 5 } .\)
Putting \(x=0, x1=B(2)+C(1)\)
\(\Rightarrow 1=2B+\frac { 3 }{ 5 } \quad \Rightarrow 2B=\frac { 2 }{ 5 } \quad \Rightarrow B=\frac { 1 }{ 5 } .\)
Comparing coeffs.of 
\(\Rightarrow 1=A+\frac { 3 }{ 5 } \Rightarrow A= \frac { 2 }{ 5 } .\)
\(From(1), \frac { { x }^{ 2 }+x+1 }{ ({ x }^{ 2 }+1)(x+2) } =\frac { \frac { 2 }{ 5 } x+\frac { 1 }{ 5 } }{ { x }^{ 2 }+1 } +\frac { \frac { 3 }{ 5 } }{ x+2 } \)
\(\therefore \int { \frac { { x }^{ 2 }+x+1 }{ ({ x }^{ 2 }+1)(x+2) } } dx=\frac { 1 }{ 5 } \int { \frac { 2x }{ { x }^{ 2 }+1 } } dx+\frac { 1 }{ 5 } \int { \frac { 1 }{ { 1 }^{ 2 }+{ x }^{ 2 } } } dx+\frac { 3 }{ 5 } \int { \frac { 1 }{ x+2 } } dx\)
\(=\frac { 1 }{ 5 } \log { \left| { x }^{ 2 }+1 \right| } +\frac { 1 }{ 5 } \tan ^{ -1 }{ x } +\frac { 3 }{ 5 } \log { \left| { x }+2 \right| } +c\)
\(=\frac { 1 }{ 5 } \log { \left| { x }^{ 2 }+1 \right| } +\frac { 3 }{ 5 } \log { \left| { x }+2 \right| } +\frac { 1 }{ 5 } \tan ^{ -1 }{ x } +c\)
\(=\log { \sqrt [ 5 ]{ ({ x }^{ 2 }+1)({ x+2) }^{ 3 } } + } \frac { 1 }{ 5 } \tan ^{ -1 }{ x } +c.\)

\(\int { { e }^{ x }\left( \frac { \sin { 4x-4 } }{ 1-\cos { 4x } } \right) } dx=\int { { e }^{ x }\left( \frac { 2\sin { 2x\cos { 2x } -4 } }{ 2\sin ^{ 2 }{ 2x } } \right) } dx\)
\(=\int { { e }^{ x }\left( \frac { 2\sin { 2x\cos { 2x } -4 } }{ 2\sin ^{ 2 }{ 2x } } -\frac { 4 }{ 2\sin ^{ 2 }{ 2x } } \right) } dx\)
\(=\int { { e }^{ x } } \left( \cot { 2x } -2{ cosec }^{ 2 }2x \right) dx\)
\(=\int { \cot { 2x } } .{ e }^{ x }dx-2\int { { cosec }^{ 2 }2x } .{ e }^{ x }dx\)
\(=\cot { 2x } .{ e }^{ x }=\int { -2{ cosec }^{ 2 }2x.{ e }^{ x }dx } -2\int { { cosec }^{ 2 }2x } .{ e }^{ x }dx\)
[Integrating first integral by Parts]
\(={ e }^{ x }\cot { 2x } +c.\)

\(\int { { \left( \frac { { x }^{ 3 }+{ 3x }+4 }{ { x }^{ 2 } } \right) } } dx\)
\(=\int { { \left( \frac { { x }^{ 3 } }{ \sqrt { x } } +3\frac { x }{ \sqrt { x } } +\frac { 4 }{ \sqrt { x } } \right) } } dx\)
\(=\int { { x }^{ \frac { 5 }{ 2 } }dx+3 } \int { { x }^{ \frac { 1 }{ 2 } }dx+4 } \int { { x }^{ \frac { 1 }{ 2 } }dx } \)
\(=\frac { { x }^{ \frac { 5 }{ 2 } +1 } }{ \frac { 5 }{ 2 } +1 } +3\frac { { x }^{ \frac { 1 }{ 2 } +1 } }{ \frac { 1 }{ 2 } +1 } +4\frac { { x }^{ -\frac { 1 }{ 2 } +1 } }{ -\frac { 1 }{ 2 } +1 } +C\)
\(=\frac { 2 }{ 7 } { x }^{ \frac { 7 }{ 2 } }+{ 2x }^{ \frac { 3 }{ 2 } }+8{ x }^{ \frac { 1 }{ 2 } }+C.\)

\(\text { Let } I=\int \frac{\sin 2 x \cos 2 x}{\sqrt{9-\cos ^{4} 2 x}} d x\)
Put cos² (2x) = t so that 4 sin 2x cos 2x dx = – dt
\(\text { Therefore } \quad \mathrm{I}=-\frac{1}{4} \int \frac{d t}{\sqrt{9-t^{2}}}=-\frac{1}{4} \sin ^{-1}\left(\frac{t}{3}\right)+\mathrm{C}=-\frac{1}{4} \sin ^{-1}\left[\frac{1}{3} \cos ^{2} 2 x\right]+\mathrm{C}\)

By definition
\(\int_{0}^{2} e^{x} d x=(2-0) \lim _{n \rightarrow \infty} \frac{1}{n}\left[e^{0}+e^{\frac{2}{n}}+e^{\frac{4}{n}}+\ldots+e^{\frac{2 n-2}{n}}\right]\)
Using the sum to n terms of a G.P., where a = 1 \(r=e^{\frac{2}{n}}, \text { we have }\)
\(\int_{0}^{2} e^{x} d x=2 \lim _{n \rightarrow \infty} \frac{1}{n}\left[\frac{e^{\frac{2 n}{n}}-1}{e^{\frac{2}{n}}-1}\right]=2 \lim _{n \rightarrow \infty} \frac{1}{n}\left[\frac{e^{2}-1}{e^{\frac{2}{n}}-1}\right]\)
\(\left.=\frac{2\left(e^{2}-1\right)}{\lim _{n \rightarrow \infty}\left[\frac{e^{\frac{2}{n}}-1}{\frac{2}{n}}\right] \cdot 2}=e^{2}-1 \quad \text { [using } \lim _{h \rightarrow 0} \frac{\left(e^{h}-1\right)}{h}=1\right] \mid\)

\(\text {(i) We have } \int { (\sin { x } } +\cos { x } )dx\)
\(=\int { \sin { xdx+ } } \int { \cos { x } } dx\)
\(=-\cos { x } +\sin { x } +c\)
\(\text {(ii) We have } \int { co\sec { x } ( } co\sec { x } +\cot { x } )dx\)
\(=\int { { co\sec { x } }^{ 2 } } x\quad dx+\int { co\sec { x } } \cot { x\quad dx } \)
\( =-\cot { x } -co\sec { x } +c\)
\(\text {(iii) We have} \int { \frac { 1-\sin { x } }{ \cos ^{ 2 }{ x } } } dx=\frac { 1 }{ \cos ^{ 2 }{ x } } dx-\int { \frac { \sin { x } }{ \cos ^{ 2 }{ x } } } dx\)
\(=\int { { sec }^{ 2 } } xdx- \int { \tan { x } \sec { x } } dx\)
\(=\tan { x } -\sec { x } +c\)

(i) We have
\(\int \frac{x^{3}-1}{x^{2}} d x=\int x d x-\int x^{-2} d x \ \text { (by Property V) }\)
\(=\left(\frac{x^{1+1}}{1+1}+C_{1}\right)-\left(\frac{x^{-2+1}}{-2+1}+C_{2}\right) ; C_{1}, C_{2} \text { are constants of integration }\)
\(=\frac{x^{2}}{2}+C_{1}-\frac{x^{-1}}{-1}-C_{2}=\frac{x^{2}}{2}+\frac{1}{x}+C_{1}-C_{2}\)
\(=\frac{x^{2}}{2}+\frac{1}{x}+\mathrm{C}, \text { where } \mathrm{C}=\mathrm{C}_{1}-\mathrm{C}_{2} \text { is another constant of integration. } \)
(ii) We have
\(\int\left(x^{\frac{2}{3}}+1\right) d x =\int x^{\frac{2}{3}} d x+\int d x \)
\(=\frac{x^{\frac{2}{3}+1}}{\frac{2}{3}+1}+x+C=\frac{3}{5} x^{\frac{5}{3}}+x+C \)
(iii) We have
\(\int\left(x^{\frac{3}{2}}+2 e^{x}-\frac{1}{x}\right) d x =\int x^{\frac{3}{2}} d x+\int 2 e^{x} d x-\int \frac{1}{x} d x \)
\(=\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}+2 e^{x}-\log |x|+\mathrm{C} \)
\(=\frac{2}{5} x^{\frac{5}{2}}+2 e^{x}-\log |x|+\mathrm{C} \) 

\(= x\ log\ (1+x^2)-2x+2\ tan^{-1}x+c\)

\(\int \frac{\sin x}{\sin x+\cos x} d x =\frac{1}{2} \int \frac{2 \sin x}{\sin x+\cos x} d x=\frac{1}{2} \int \frac{(\sin x+\cos x)+(\sin x-\cos x)}{\sin x+\cos x} d x \)
\(=\frac{1}{2}\left[\int 1 . d x+\int \frac{\sin x-\cos x}{\sin x+\cos x} d x\right] \mid \begin{array}{l} \text { Let } \sin x+\cos x=t \\ \Rightarrow(\cos x-\sin x) d x=d t \end{array}\)
\(=\frac{1}{2}\left[x+\int-\frac{1}{t} d t\right]=\frac{1}{2}[x-\log |\sin x+\cos x|]+C \)

\(\int \frac{e^{x}}{\sqrt{5-4 e^{x}-e^{2 x}}} =\int \frac{1}{\sqrt{5-4 t-t^{2}}} d t=\int \frac{1}{\sqrt{9-(t+2)^{2}}} d t \)
\(=\sin ^{-1}\left(\frac{t+2}{3}\right)+C=\sin ^{-1}\left(\frac{e^{x}+2}{3}\right)+C \)