How do you integrate #∫(x^3ex^2)/(x^2 +1)^2dx#?

#∫(x^3e^(x^2))/(x^2 +1)^2dx # ?

2 Answers
Jul 12, 2018

#int(x^3e^(x^2))/(x^2+1)^2dx=e^(x^2)/(2(x^2+1))+c#

Explanation:

Here,

#I=int(x^3e^(x^2))/(x^2+1)^2dx#

#=int(x^2e^(x^2))/(x^2+1)^2*xdx#

Subst . #color(violet)(x^2=u=>2xdx=du=>xdx=1/2du#

So,

#I=int(ue^u)/(u+1)^2*1/2du#

#=1/2int((u+1-1)e^u)/(u+1)^2du#

#=1/2int{((u+1))/(u+1)^2-1/(u+1)^2}e^udu#

#=1/2{color(blue)(int1/(u+1)e^udu)-color(red) (int1/(u+1)^2e^udu)}#

Using Integration by parts: in the first integral

#I#=#1/2{color(blue)([1/(u+1)inte^u-int(-1)/(u+1)^2e^udu])- color(red)(int1/(u+1)^2e^udu)}#

#I#=#1/2{color(brown)(1/(u+1)e^u+color(blue) (int(1)/(u+1)^2e^udu)-color(red)(int1/(u+1)^2e^udu)}#

#I=1/2 *color(brown)(1/(u+1)e^u)+c#

Subst. back #color(violet)( u=x^2# we get

#I=1/2(1/(x^2+1))e^(x^2)+c#

Hence ,

#I=e^(x^2)/(2(x^2+1))+c#
.....................................................................................

Note: Change in the question is as below.

#x^3ex^2tox^3e^(x^2)#

Jul 12, 2018

The answer is #=(e^(x^2))/(2(x^2+1))+C#

Explanation:

Perform the substitution

#u=x^2#, #=>#, #du=2xdx#

The integral is

#I=int(x^3e^(x^2)dx)/(x^2+1)^2#

#=1/2int(ue^udu)/(u+1)^2#

Perform an integration by parts

#intwv'dx=wv-intw'vdx#

#w=ue^u#, #=>#, #w'=ue^u +e^u=e^u(u+1)#

#v'=1/(u+1)^2#, #=>#, #v=-1/(u+1)#

Therefore,

#I=-1/2(ue^u)/(u+1)+1/2inte^udu#

#=e^u/2-(ue^u)/(2(u+1))#

#=e^(x^2)/2-(x^2e^(x^2))/(2(x^2+1))+C#

#=1/2e^(x^2)((x^2+1-x^2))/(x^2+1)+C#

#=(e^(x^2))/(2(x^2+1))+C#