How do you find the antiderivative of $int x^3/(4+x^2) dx$ ?

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Answer 1 · 2016-12-02T22:28:08

$I=intx^3/(4+x^2)dx$

Let $u=4+x^2$ , implying that $du=2xdx$ . Also note that $x^2=u-4$ . Rearranging the integral:

$I=1/2int(x^2(2xdx))/(4+x^2)=1/2int(u-4)/udu=1/2int(1-4/u)du$

Splitting up the integral:

$I=1/2intdu-2int1/udu=1/2u-2lnabsu=1/2(4+x^2)-2lnabs(4+x^2)+C$

Letting the constant from $1/2(4+x^2)$ absorb into $C$ :

$I=1/2x^2-2ln(x^2+4)+C$

How do you find the antiderivative of int x^3/(4+x^2) dxint x^3/(4+x^2) dx?