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

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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,

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

$=> f(x) + c$

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

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