Find integral of #(((x^2)+1).e^x)/(x+1)^2 # With respect to #x#?

Question

3385 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-07T07:49:11

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

#=>inte^x[(((x^2-1)+2))/(x+1)^2] dx#

#=>inte^x[((x-1)(x+1)+2)/(x+1)^2] dx#

#=>inte^x[(x-1)/(x+1)]dx+2inte^x[1/(x+1)^2]dx#

Notice,
Considering #f(x) = [(x-1)/(x+1)]#

#f'(x) = ((x+1)(x-1)'-(x-1)(x+1)')/(x+1)^2#

#=> ((x+1)-(x-1))/(x+1)^2 => 2/(x+1)^2#

So the integral is now of the form,

#=>inte^x[f(x)]dx+inte^x[f'(x)]dx#

#=> f(x) + c#

#=> e^x[(x-1)/(x+1)] + c#