Let x = tan \(\theta \Rightarrow\)dx = sec²\(\theta d \theta\)

\(\therefore I=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \frac { log(1+tan\theta ) }{ { sec }^{ 2 }\theta } } { sec }^{ 2 }\theta d\theta \)
\(=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ log(1+tan\theta )d\theta } \quad ...(1)\)
Using property,
\(I=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ log(1+tan(\frac { \pi }{ 4 } -\theta ))d\theta } \)
\(I=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ log\left( \frac { 1+tan\theta +1-tan\theta }{ 1+tan\theta } \right) } d\theta \)
\(I=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ log\left( \frac { 2 }{ 1+tan\theta } \right) } d\theta \quad ..(2)\)
(1)+(2)
\(2I=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ log(1+tan\theta )d\theta } +\int _{ 0 }^{ \frac { \pi }{ 4 } }{ log\left( \frac { 2 }{ 1+tan\theta } \right) } d\theta \)

\(=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ log2d\theta =log2{ [\theta ] }_{ 0 }^{ \frac { \pi }{ 4 } } } \)
\(2I=log2(\frac { \pi }{ 4 } -0)=\frac { \pi }{ 4 } log2\)
\(\therefore I=\frac { \pi }{ 8 } log2\)

Let I = \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { cos\theta }{ (1+sin\theta )(2+sin\theta ) } } d\theta \)
Put u = 1 + sin\(\theta\)
Then, du = cos\(\theta\) d\(\theta\)
When \(\theta\) = 0, u = 1 
When \(\theta =\frac{\pi}{2}, u=2\)
\(\therefore I=\int _{ 1 }^{ 2 }{ \frac { du }{ u(1+u) } } =\int _{ 1 }^{ 2 }{ \frac { (1+u)-u }{ u(1+u) } du } =\int _{ 1 }^{ 2 }{ \left( \frac { 1 }{ u } -\frac { 1 }{ 1+u } \right) du=[logu-log(1+u)]_{ 1 }^{ 2 } } \)
\(=(log2-log3)-(log1-log2)=2log2-log3=log\frac { 4 }{ 3 } .\)

Let I = \(\int _{ 1 }^{ 2 }{ \frac { x }{ (x+1)(x+2) } dx } \)
\(I=\int _{ 1 }^{ 2 }{ \left[ \frac { -1 }{ (x+1) } +\frac { 2 }{ x+2 } \right] } dx\) (Using partial fractions)
\(={ [-log(x+1)+2log(x+2)] }_{ 1 }^{ 2 }\)
\(=log{ \left[ \frac { { (x+2) }^{ 2 } }{ x+1 } \right] }_{ 1 }^{ 2 }\)
\(=log\frac { 16 }{ 3 } -log\frac { 9 }{ 2 } \)
\(=log\frac { 32 }{ 27 } \)

Here a = 0, b = 1,f (x) = 5x + 4
\(\therefore f(a+(b-a)\frac { r }{ n } )=f\left( 0+1(\frac { r }{ n } ) \right) =f(\frac { r }{ n } )=5(\frac { r }{ n } )+4\)
\(\int _{ a }^{ b }{ f(x)dx } =\underset { n\rightarrow \infty }{ lim } \frac { b-a }{ n } \sum _{ r=1 }^{ n }{ f(a+(b-a)\frac { r }{ n } } \)
\(\therefore \int _{ 0 }^{ 1 }{ (5x+4)dx } =\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } \sum _{ r=1 }^{ n }{ \left( \frac { 5r }{ n } +4 \right) } \)
\(=\underset { n\rightarrow \infty }{ lim } \left[ \frac { 1 }{ n } \sum _{ r=1 }^{ n }{ \frac { 5r }{ n } +\frac { 1 }{ n } \sum _{ r=1 }^{ n }{ 4 } } \right] \)
\(=\underset { n\rightarrow \infty }{ lim } \left[ \frac { 1 }{ n } .\frac { 5 }{ n } .(1+2+3+...+n)+\frac { 1 }{ n } .4n \right] \)
\(=\underset { n\rightarrow \infty }{ lim } \left[ \frac { 5 }{ { n }^{ 2 } } \frac { n(n+1) }{ 2 } +4 \right] \)
\([\because \sum { r=\frac { n(n+1) }{ 2 } ;\sum { { r }^{ 2 }=\frac { n(n+1)(2n+1) }{ 6 } } } ]\)
\(=\underset { n\rightarrow \infty }{ lim } \frac { 5 }{ { n }^{ 2 } } { n }^{ 2 }\frac { (1+\frac { 1 }{ n } ) }{ 2 } +4\)
\(=\frac { 5 }{ 2 } (1+0)+4=\frac { 5 }{ 2 } +4\)
\([wehen\quad n\rightarrow \infty ,1/n\quad \rightarrow 0]\)
\(=\frac { 5+8 }{ 2 } =\frac { 13 }{ 2 } \)
\(\therefore \int _{ 0 }^{ 1 }{ (5x+4)dx } =\frac { 13 }{ 2 } \)

Given that I =\(\int ^\frac {\pi}{2}_{0} \)( sin²x + cos⁴x)dx =\(\int ^\frac {\pi}{2}_{0} \) sin²x dx+\(\int ^\frac {\pi}{2}_{0} \)cos⁴x dx\(\frac {1}{2} \times \frac {\pi}{2} + \frac {3}{4} \times \frac {1}{2} \times \frac {\pi}{2} = \frac {7\pi}{16} \)

\(\int _{ 3 }^{ 4 }{ \frac { dx }{ { x }^{ 2 }-4 } } \)
\(\int _{ 3 }^{ 4 }{ \frac { dx }{ { x }^{ 2 }-4 } } =\int _{ 3 }^{ 4 }{ \frac { dx }{ { x }^{ 2 }-{ 2 }^{ 2 } } } \)
\(\left[ \because \int { \frac { dx }{ { x }^{ 2 }-{ a }^{ 2 } } } =\frac { 1 }{ 2a } log\left| \frac { x-a }{ x+a } \right| +c\right] \)
\(=\frac { 1 }{ 4 } \left[ log\left( \frac { 4-2 }{ 4+2 } \right) -log\left( \frac { 3-2 }{ 3+2 } \right) \right] \)
\(=\frac { 1 }{ 4 } log\left[ \left( \frac { 2 }{ 6 } \right) - log \ \frac { 1 }{ 5 } \right] \\ =\frac { 1 }{ 4 } log\left( \frac { 1 }{ 3 } \times 5 \right) \)
\(=\frac { 1 }{ 4 } log\left( \frac { 5 }{ 3 } \right) \)

Let f(x) = e^-|-x| = e^-|x| = f(x)
So f (x) is an even function.
Hence, \(\int ^{log 2}_{-log 2} e ^{-|x|}\)dx = 2\(\int ^{log 2}_{0} e ^{-|x|}\)dx = 2\(\int ^{log 2}_{0} e ^{-x}\) dx
= 2(-e^-x)\(^{log2}_{0}\) = 2 (-e^-log2 + e⁰) = 2 \((-e ^{log \frac{1}{2}} + 1)\)
= 2\((-\frac {1}{2}+1)=1\).

Let f (x) = x cos x
Then f (−x) = (−x) cos(−x) = −x cos x = − f (x).
So f (x) = x cos x is an odd function.
Hence, applying the property, for odd function f(x), \(\int _{ -a }^{ a }{ f(x)dx=0 } \)
\(\therefore\) we get \(\int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ xcosx } dx=0\)

Taking u = 1−x² and v = e^{−\(\lambda\)x}, and applying the Bernoulli’s formula, we get
\(I=\int _{ -1 }^{ 1 }{ { e }^{ -\lambda x } } (1-{ x }^{ 2 })dx={ \left[ (1-{ x }^{ 2 })\left( \frac { { e }^{ -\lambda x } }{ -\lambda } \right) -(-2x)\left( \frac { { e }^{ -\lambda x } }{ { \lambda }^{ 2 } } \right) +(-2)\left( \frac { { e }^{ -\lambda x } }{ -{ \lambda }^{ 3 } } \right) \right] }_{ -1 }^{ 1 }\)
\(=2\left( \frac { { e }^{ -\lambda } }{ { \lambda }^{ 2 } } \right) +2\left( \frac { { e }^{ -\lambda } }{ { \lambda }^{ 3 } } \right) +2\left( \frac { { e }^{ \lambda } }{ { \lambda }^{ 2 } } \right) -2\left( \frac { { e }^{ \lambda } }{ { \lambda }^{ 3 } } \right) \)
\(=\frac { 2 }{ { \lambda }^{ 2 } } ({ e }^{ \lambda }+{ e }^{ -\lambda })-\frac { 2 }{ { \lambda }^{ 3 } } ({ e }^{ \lambda }-{ e }^{ -\lambda })\)

Let \(f(x)=xcos\left( \frac { { e }^{ x }-1 }{ { e }^{ x }+1 } \right) \)
\(f(-x)=-x\quad cos\left( \frac { { e }^{ x }-1 }{ { e }^{ x }+1 } \right) \)
\(=-x\quad cos\left( \frac { { e }^{ \frac { 1 }{ x } }-1 }{ { e }^{ \frac { 1 }{ x } }+1 } \right) \)
\(=-x\quad cos\left( \frac { 1-{ e }^{ x } }{ 1+{ e }^{ x } } \right) \)
\(=-x\quad cos\left( -\left( \frac { { e }^{ x }-1 }{ { e }^{ x }+1 } \right) \right) \)
\(=-xcos\left( \frac { { e }^{ x }-1 }{ { e }^{ x }+1 } \right) \)
\([\because cos(-\theta )=cos\theta ]\)
= -f(x)
\(\therefore\) f(x) is an odd function
\(\therefore \int _{ -5 }^{ 5 }{ xcos\left( \frac { { e }^{ x }-1 }{ { e }^{ x }+1 } \right) } dx=0\)

I = \(\int _{ 0 }^{ 1 }{ \frac { 1-{ x }^{ 2 } }{ { (1+{ x }^{ 2 }) }^{ 2 } } } dx\)
[Dividing the numerator and denominator by x]
\(I=\int _{ 0 }^{ 1 }{ \frac { \frac { 1 }{ { x }^{ 2 } } -\frac { { x }^{ 2 } }{ { x }^{ 2 } } }{ 0\left( \frac { 1 }{ x } +\frac { { x }^{ 2 } }{ { x }^{ 2 } } \right) } } dx=\int _{ 0 }^{ 1 }{ \frac { { x }^{ \frac { 1 }{ 2 } -1 } }{ { (\frac { 1 }{ x } +x) }^{ 2 } } } dx\)
\(put\ x+\frac { 1 }{ x } =t\Rightarrow (1-\frac { 1 }{ { x }^{ 2 } } )dx=dt\)
\(\Rightarrow \left( \frac { 1 }{ { x }^{ 2 } } -1 \right) dx=-dt\)

\(\therefore I=\int _{ \infty }^{ 2 }{ -\frac { dt }{ { t }^{ 2 } } } =\int _{ 2 }^{ \infty }{ \frac { dt }{ { t }^{ 2 } } } \)
\(=\int _{ 2 }^{ \infty }{ { t }^{ -2 }dt } ={ \left[ \frac { { t }^{ -1 } }{ -1 } \right] }_{ 2 }^{ \infty }={ \left[ -\frac { 1 }{ t } \right] }_{ 2 }^{ \infty }\)
\(=-\frac { 1 }{ \infty } +\frac { 1 }{ 2 } =0+\frac { 1 }{ 2 } \)
\(=\frac { 1 }{ 2 } \therefore \int _{ 0 }^{ 1 }{ \frac { 1-{ x }^{ 2 } }{ { (1+{ x }^{ 2 }) }^{ 2 } } dx=\frac { 1 }{ 2 } } \)

Let I = \(\int ^{a}_{0} \frac{f(x)}{f(x)+f(a-x)}\)    ------- (1)
Applying the formula \(\int ^{a}_{0}\) f(x) dx = \(\int ^{a}_{0}\)f(a-x)dx  in equation (1), we get
 I =  \(\int ^{a}_{0} \frac{f{(a-x)}}{f{(a-x)}+f(a-(a-x))}\) dx
= \(\int ^{a}_{0} \frac{f{(a-x)}}{f{(x)}+f(a-x)}\) dx----- (2)
Adding equations (1) and (2), we get
2I =  \(\int ^{a}_{0} \frac{f{(x)}}{f{(x)}+f(a-x)}\) dx + \(\int ^{a}_{0} \frac{f{(a-x)}}{f{(x)}+f(a-x)}\) dx
= \(\int ^{a}_{0} \frac{f(x) +f{(a-x)}}{f{(x)}+f(a-x)}\) dx
= \(\int ^{a}_{0} dx = a\)
Hence, we get I = \(\frac{a}{2}\)