\(\left[ \because \int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -ax }dx=\frac { n! }{ { a }^{ n+1 } } } \right] \)
இங்கு n = 4,a = 1
\(\int _{ 0 }^{ \infty }{ x^{ 4 }{ e }^{ -x } } dx =\frac { 4! }{ { 1 }^{ 3 } }= 4! \)

\(\Gamma \)(1) = 0! = 1

 \(\Gamma \)8 = 7! = 5040

\(\int _{ 0 }^{ 4 }{ \left( \sqrt { x } +\frac { 1 }{ \sqrt { x } } \right) } \)dx \(=\int _{ 0 }^{ 4 }{ \left( { x ^ {1/2}} +{ x^{ -1/2} } \right) } dx \)
\(= \left [ \frac {x^{3/2}}{3/2} + \frac{x^{1/2}}{1/2} \right]^4_0\)
\(=2 \left [ \frac {4^{3/2}}{3} + 4^{1/2}\right]\)
\(=2 \left [ \frac {8}{3} + 2\right] = \frac{28}{3}\)

 \(\int _{ 0 }^{ 1 }{ (a-x)^{ 2 } } \) f(x )dx = \(\int _{ 0 }^{ 1 }{ (x^2 -2a-x)^{ 2 } } f(x)dx\)
= \(a^2 \int f(x) dx-2a \int _{ 0 }^{ 1 }x f(x) dx + \int _{ 0 }^{ 1 }x^2 f(x) dx\)
= a² - 2a² + a² = 0

\(\int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \cos x } dx=2\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \cos x } dx\)
( cos x ஒரு இரட்டைச் சார்பு)
\(= [\sin x]^{\pi/2}_ 0 = 2[1-0]= 2\)

t = x⁴ என்க.⇒ dt = 4x³ dx ⇒ dt/4 = x³ dx 
இங்கு x = 1 எனில் t = 1
x = -1 எனில் t = 1
\(\int _{ 0 }^{ 1 }{ { x }^{ 3 }{ e }^{ { x }^{ 4 } } } dx =\frac {1}{4}\int _{ 1 }^{ 1 }e ^{t} dt = 0\)

\(\int _{ 2 }^{ 4 }{ \frac { dx }{ x } } = { { \left[ \log x \right] }^{ 4 }_{ 2 } }\)
= log 4 - log 2 = log 4/2 = log 2

t = x² + 3x + 2 என்க.⇒ dt = (2x+ 3) dx
\(\int { \frac { 2x+3 }{ \sqrt { { x }^{ 2 }+3x+2 } } } \)dx \(=\int \frac {dt }{\sqrt t} = \int t ^{- \frac {1}{2}}dt = \frac {t {\frac {1}{2}}}{\frac {1}{2}}\)
= \(2 \sqrt{t } + c =2\sqrt { { x }^{ 2 }+3x+2 } +c\)

t = 4 + x⁴ என்க.⇒ dt = x⁴ dx
\(\int { \frac { { 2x }^{ 3 } }{ 4+{ x }^{ 4 } } } \)dx \(= \int \frac {1}{2}{ \frac { { 4x }^{ 3 } }{ 4+{ x }^{ 4 } } } dx = \frac {1}{2} \int \frac {dt }{t }\)
\(=\frac { 1 }{ 2 } \log \left| t \right| +c\)
\(=\frac { 1 }{ 2 } \log \left| 4+{ x }^{ 4 } \right| +c\)

\(∵ \int \frac {f '(x)}{f (x )} dx = \log |f (x )|+c \)
இங்கு f (x ) = e^x + 1 ; f'(x ) = e^x 
∴ \(\int { \frac { { e }^{ x } }{ { e }^{ x }+1 } } \)dx = log|e^x+1|+c

t = e^x என்க. ⇒ dt = e^x dx ⇒ \(dx = \frac {dt}{e^x} = \frac {dt}{t }\)
\(\int \sqrt {e^ x }dx = \int \sqrt {t } \frac {dt}{t} = \int \frac {dt}{ \sqrt t} \)
\(= \int t^ {\frac {-1}{2}}dt = 2\sqrt {t} \)
= 2\(\sqrt { { e }^{ x } } +c\)

\(\\ \int { \frac { sin5x-sinx }{ cos3x } } \)dx = \(\int { \frac { 2\sin \frac {5x-x }{2}\cos \frac {5x+x }{2} }{ \cos3x } } \)
=2 ∫sin 2x dx = \(2\left [\frac {−\cos 2x}{2} \right] +c \)
= −cos2x + c 

\(\int { \frac { \sin2x }{ 2\sin x } } \)dx = \(\int { \frac { 2\sin \cos x }{ 2\sin x } } dx \)
= ∫cos x = ∫sin x + c

\(\int { \frac { 1 }{ { x }^{ 3 } } dx } =\int { { { x }^{ 3 } } dx } = \frac { x^ {-2} }{-2} + c = \frac { -1 }{ 2x^{ 2 } } +x\)