Find integral of (((x^2)+1).e^x)/(x+1)^2 ((x2)+1).ex(x+1)2 With respect to xx?

1 Answer
Apr 7, 2018

I=int((x^2+1)*e^x)/(x+1)^2 dxI=(x2+1)ex(x+1)2dx

=>inte^x[(((x^2-1)+2))/(x+1)^2] dxex[((x21)+2)(x+1)2]dx

=>inte^x[((x-1)(x+1)+2)/(x+1)^2] dxex[(x1)(x+1)+2(x+1)2]dx

=>inte^x[(x-1)/(x+1)]dx+2inte^x[1/(x+1)^2]dxex[x1x+1]dx+2ex[1(x+1)2]dx

Notice,
Considering f(x) = [(x-1)/(x+1)]f(x)=[x1x+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