\(\Gamma \)(n) =\(\int _{ 0 }^{ \infty }{ { t }^{ n-1 }e^{ -t } } \) என்பதை நாம் அறிவோம்.
இதில் t=x² என dt=2xdx ஆகும்.
∴ \(\Gamma \)(n) =\(\int _{ 0 }^{ \infty }{ { e }^{ { -x }^{ 1 } }({ x }^{ 2 })^{ n-1 } } 2xdx\)
=\(\int _{ 0 }^{ \infty }{ { e }^{ { -x }^{ 1 } }x^{ 2x-2 } } 2xdx\)
=\(2\int _{ 0 }^{ \infty }{ { e }^{ { -x }^{ 1 } }x^{ 2x-1 } } dx\)
n=\(\frac{1}{2}\) என, நாம் பெறுவது
\(\Gamma \left( \frac { 1 }{ 2 } \right) =2\int _{ 0 }^{ \infty }{ { e }^{ { -x }^{ 1 } } } dx\)
⇒ \(\sqrt { \pi } \int _{ 0 }^{ \infty }{ e^{ { -x }^{ 2 } } } dx\)
∴ \(\int _{ 0 }^{ \infty }{ e^{ { -x }^{ 2 } } } dx=\frac { \sqrt { \pi } }{ 2 } \)

\(\int { \sqrt { x^{ 2 }+5 } } dx=\int { \sqrt { { x }^{ 2 }+(\sqrt { 5 } )^{ 2 } } } dx\)
\(=\frac { x }{ 2 } \sqrt { { x }^{ 2 }+(\sqrt { 5 } )^{ 2 } } +\frac { (\sqrt { 5 } )^{ 2 } }{ 2 } \log|x+\sqrt { { x }^{ 2 }+\left( \sqrt { 5 } \right) ^{ 2 } } |+c\)
\(=\frac { x }{ 2 } \sqrt { { x }^{ 2 }+5 } +\frac { 5 }{ 2 } \log|x+\sqrt { { x }^{ 2 }+5 } |+c\)

\(\int \frac { { x }^{ 3 } }{ x+2 } dx\)

\(x^{ 3 }=(x+2)\left( x^{ 2 }-2x+4 \right) -8\)
\(\int \frac { x^{ 3 } }{ x+2 } dx=\int \left( x^{ 2 }-2x+4 \right) dx-8\int \frac { dx }{ x+2 }\)
\(=\frac { x^{ 3 } }{ 3 } -\frac { 2x^{ 2 } }{ 2 } +4x-8\log |x+2|+c\)
\(=\frac { x^{ 3 } }{ 3 } -x^{ 2 }+4x-8\log |x+2|+c\)

\(\int (3+x)(2-5x)dx\quad =\int \left( 6-15x+2x-5x^{ 2 } \right) dx\)
\(=\int \left( 6-13x-5x^{ 2 } \right) dx\)
\(=6x-\frac { 13x^{ 2 } }{ 2 } -\frac { 5x^{ 3 } }{ 3 } +c\)

காரணிகளாக மாற்றியமைக்க, 2x²-14x+24=(x-3)(2x-8)
\(\int { \frac { { 2x }^{ 2 }-14x+24 }{ x-3 } } \)=dx =\(\int { \frac { \left( x-3 \right) \left( 2x-8 \right) }{ x-3 } } dx\)
=\(\int { \left( 2x-8 \right) } dx\)
= x² - 8x + c

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

\(\int { \frac { 2 }{ 3x+5 } dx } =2\int { \frac { 1 }{ 3x+5 }} \)dx
\(=\frac { { 2 } }{ 3 } \log\left| 3x+5 \right| +c\)

\(\int { \frac { { 3x }^{ 2 }+6x+1 }{ \left( x+3 \right) \left( { x }^{ 2 }+1 \right) } } dx =\int { \left[ \frac { 1 }{ \left( x+3 \right) } \frac { 2x }{ \left( { x }^{ 2 }+1 \right) } \right] } dx\)
\(\int { \frac { dx }{ \left( x+3 \right) } +\int { \frac { 2x }{ \left( { x }^{ 2 }+1 \right) } } } dx\)
\(=\log\left| x+3 \right| +\log\left| { x }^{ 2 }+1 \right| +c\)
\(=\log\left| \left( x+3 \right) \left( { x }^{ 2 }+1 \right) \right| +c\)
\(=\log\left| { x }^{ 3 }+{ 3x }^{ 2 }+x+3 \right| +c\)
பகுதி பின்னமாக்கல் முறையில்
\(\frac { { 3x }^{ 2 }+6x+1 }{ \left( x+3 \right) \left( { x }^{ 2 }+1 \right) } =\frac { A }{ \left( x+3 \right) } +\frac { Bx+C }{ \left( { x }^{ 2 }+1 \right) }\)
\( \Rightarrow \frac { { 3x }^{ 2 }+6x+1 }{ \left( x+3 \right) \left( { x }^{ 2 }+1 \right) } =\frac { 1 }{ \left( x+3 \right) } +\frac { 2x }{ \left( { x }^{ 2 }+1 \right) } \).

\(\int { \frac { dx }{ 2+x-x^{ 2 } } } \)=\(\int { \frac { dx }{ 2\left( \frac { 3 }{ 2 } \right) ^{ 2 }-\left( x-\frac { 1 }{ 2 } \right) ^{ 2 } } } \)
=\( \frac { 1 }{ 2\left( \frac { 3 }{ 2 } \right) } \log\left| \frac { \frac { 3 }{ 2 } +(x-\frac { 1 }{ 2 } ) }{ \frac { 3 }{ 2 } (x-\frac { 1 }{ 2 } ) } \right| \)+c
=\(\frac { 1 }{ 3 } \log\left| \frac { 2+2x }{ 4-2x } \right| \)+c
=\(\frac { 1 }{ 3 } \log\left| \frac { 1+x }{ 2-x } \right| \)+c
வர்க்க நிறைவாக்கல் முறையில்,
2+x-x²=2-[x²-x]
= 2-\(\left[ (x-\frac { 1 }{ 2 } )^{ 2 }-\frac { 1 }{ 4 } \right] \)
=\(\left( \frac { 3 }{ 2 } \right) ^{ 2 }-\left( x-\frac { 1 }{ 2 } \right) ^{ 2 }\)

\(\int { { e }^{ x }\left( { x }^{ 2 }+2x \right) dx } =\int { { e }^{ x }\left[ f(x)+f'(x) \right] } dx\)
= e^x f (x) + c
= e^x x² + c
f (x) = x^{2 
\(\therefore\)} f '(x) = 2x

\(\int \left( \frac { e^{ 3x }+e^{ 5x } }{ e^{ x }+e^{ -x } } \right) dx\)\(=\int \frac { e^{ 3x }\left( 1+{ e }^{ 2x } \right) }{ e^{ -x }\left( 1+{ e }^{ 2x } \right) } dx\)
\(=\int e^{ 4x }dx=\frac { e^{ 4x } }{ 4 } +c\)

\(\int { { 3 }^{ 2x+3 } } \)dx=\(\int { { 3 }^{ 2x } } \).3³dx
= 3³\(\int { { 3 }^{ 2x } } \)dx
= 27\(\frac { 3^{ 2x } }{ 2\log3 } \)+c

f'(x)=\(\frac { 1 }{ x } \) 
f(x) = \(\int \frac { dx }{ x } \)
f(x)= log|x| +c
f(1) =\(\frac { \pi }{ 4 } \)
\(\frac { \pi }{ 4 } \)=log 1 +c
c=\(\frac { \pi }{ 4 } \)  (ஃlog 1 =0) 
f(x) = log|x| + \(\frac { \pi }{ 4 } \)

\(\int \frac { 1 }{ \sqrt { x+1 } +\sqrt { x-1 } } dx\)
\(\int \frac { 1 }{ \sqrt { x+1 } } +\sqrt { x-1 } \times \frac { \sqrt { x+1 } -\sqrt { x-1 } }{ \sqrt { x+1 } =\sqrt { x-1 } } dx\)
\(=\int \frac { \sqrt { x+1 } -\sqrt { x-1 } }{ x+1-(x-1) } dx\)
\(=\int { \frac { (x+1)^{ \frac { 1 }{ 2 } }-(x-1)^{ \frac { 1 }{ 2 } } }{ 2 } } \)
\(=\frac { 1 }{ 2 } \left[ \frac { (x+1)^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } -\frac { (x-1)^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } \right] \)
\(=\frac { 1 }{ 3 } \left[ (x+1)^{ \frac { 3 }{ 2 } }-(x-1)^{ \frac { 3 }{ 2 } } \right] +{ c }\)

\(\int _{ 0 }^{ \frac { \pi }{ 3 } }{ \tan x } dx = [ \log \sec x ]^ \frac { \pi }{ 3 }_ 0\)
= log sec π/3 - log sec 0
= log2 - log1 = log 2

\(\int _{ 0 }^{ 1 }{ \sqrt { { x }^{ 4 }(1-x)^{ 2 } } }\)dx = \(\int _{ 0 }^{ 1 }{ \sqrt { { x }^{ 2 }(1-x)^{ 2 } } }\)
\(=\int _{ 0 }^{ 1 }{ \sqrt { { x }^{ 2 }-x^{ 3 } } }dx = \left[\frac{x^{3}}{3} \frac{x^4}{4}\right]^1_0\)
\(= \frac{1}{3} - \frac{1}{4} = \frac{4-3}{12} = \frac{1}{12}\)

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

t = log x என்க.\(\Rightarrow dt = \frac { 1 }{ x } dx\)
\(\int \frac { \log x }{ x } dx =\int tdt = { \frac { { t }^{ 2 } }{ 2 } } +c = \frac {1}{2}(\log x ) ^ 2+c \)

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