What is the antiderivative of $x^3/(1+x^2)$ ?

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Answer 1 · 2016-07-07T16:51:03

$1/2{x^2+1-ln(x^2+1)}+C.$

OR

$1/2{x^2-ln(x^2+1)}+C_1,$ where $C_1=C+1/2.$

Let $I=intx^3/(1+x^2)dx$

We take substn. $x^2+1=t$ , so that, $2xdx=dt.$

Also, $x^2+1=t rArr x^2=t-1$

Now, $I=intx^3/(1+x^2)dx=1/2int(x^2*2x)/(1+x^2)dx=1/2int(t-1)/tdt$
$=1/2int{t/t-1/t}dt=1/2int{1-1/t}dt=1/2{t-lnt}=1/2{x^2+1-ln(x^2+1)}+C.$

$I$ is also $=1/2x^2+1/2-1/2ln(x^2+1)+C$
$=1/2{x^2-ln(x^2+1)}+C_1,$ where $C_1=C+1/2$

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What is the antiderivative of x^3/(1+x^2)x^3/(1+x^2)?