Question

Intergration of
$\bf \int( {x}^{2} {e}^{ {x}^{5} } )dx$
​

Answer 1

Topic :-

Calculus - II

To Solve :-

$\displaystyle \int x^2e^{x^{5}}dx$

Solution :-

$\displaystyle \int x^2e^{x^{5}}dx$

$Substitute,\:u=x^3$

$Differentiating\:\:both\:\:sides,$

$du=3x^2dx$

$dx=\dfrac{1}{3x^2}du$

Now, we can write it as,

$\dfrac{1}{3}\displaystyle \int e^{u^{5/3}}du$

This is a special integral (incomplete gamma function)

So, it becomes,

$\dfrac{\Gamma \left ( \dfrac{3}{5},-u^{5/3} \right )}{5}$

Put back the values of 'u'.

$\dfrac{\Gamma \left ( \dfrac{3}{5},-x^{5} \right )}{5}$

Answer :-

$\dfrac{\Gamma \left ( \dfrac{3}{5},-x^{5} \right )}{5}+C$

Extra Information :-

Incomplete Gamma Function

The upper and lower incomplete gamma functions are types of special functions which arise as solutions to various mathematical problems such as certain integrals.

$\Gamma(z,x)=x^ze^{-x}\displaystyle \sum_{n=0}^{\infty}\dfrac{L_n^{(z)}(x)}{n+1}$,

which converges for R(z) > -1 and x > 0.

Answer 2

$\huge{\fbox{\fcolorbox{cyan}{pink}{✿AnsWer}}}$

$: \longmapsto \: {\boxed {\red{ \frac{1}{2} {x}^{4} {e}^{ {x}^{2} } - {x}^{2} {e}^{ {x}^{2} } + {e}^{ {x}^{2} } + c}}}$

◑ TO FIND:-

Integration of ---

$\int \frac{1}{2} {x}^{4} {e}^{ {x}^{2} } dx$

______________________________________________★

◑ EXPLANATION:-

$: \longrightarrow \: \int {x}^{5} . {e}^{ {x}^{2} } dx$

$: \longrightarrow \: \frac{1}{2} \int {x}^{4} {e}^{ {x}^{2} }. \: 2xdx$

Integration by substitution:-

$: \longrightarrow \: {x}^{2} = u$

$: \longrightarrow \: d( {x}^{2} ) = du$

$: \longrightarrow \frac{1}{2} \int {x}^{4} {e}^{ {x}^{2} } d {x}^{2}$

$: \longrightarrow \frac{1}{2} \int {u}^{2} {e}^{u} du$

Integration by parts:-

$: \longrightarrow \boxed{ \frac{1}{2} \int {u}^{2} d( {e}^{u})}$

$: \longrightarrow \frac{1}{2} ( {u}^{2} {e}^{u} - \int2u {e}^{u} du$

Using Integration by parts again:-

$: \longrightarrow \frac{1}{2}( {u}^{2} {e}^{u} - 2 \int \: ud( {e}^{u} ) )$

$: \longrightarrow \frac{1}{2} ( {u}^{2} {e}^{u} - 2u {e}^{u} - ( - 2 \int {e}^{u} du))$

$: \longrightarrow \frac{1}{2} {u}^{2} {e}^{u} - u {e}^{u} + {e}^{u} + c$

Reverse The substitution:-

$: \longmapsto \: {\boxed {\purple{ \frac{1}{2} {x}^{4} {e}^{ {x}^{2} } - {x}^{2} {e}^{ {x}^{2} } + {e}^{ {x}^{2} } + c}}}$

HOPE IT HELPS