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

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

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Answer 1 · 2016-12-02T22:28:08

$I = \int \frac{x^{3}}{4 + x^{2}} d x$ $I=intx^3/(4+x^2)dx$

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

$I = \frac{1}{2} \int \frac{x^{2} (2 x d x)}{4 + x^{2}} = \frac{1}{2} \int \frac{u - 4}{u} d u = \frac{1}{2} \int (1 - \frac{4}{u}) d u$ $I=1/2int(x^2(2xdx))/(4+x^2)=1/2int(u-4)/udu=1/2int(1-4/u)du$

Splitting up the integral:

$I = \frac{1}{2} \int d u - 2 \int \frac{1}{u} d u = \frac{1}{2} u - 2 ln | u | = \frac{1}{2} (4 + x^{2}) - 2 ln ∣ ∣ 4 + x^{2} ∣ ∣ + C$ $I=1/2intdu-2int1/udu=1/2u-2lnabsu=1/2(4+x^2)-2lnabs(4+x^2)+C$

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

$I = \frac{1}{2} x^{2} - 2 ln (x^{2} + 4) + C$ $I=1/2x^2-2ln(x^2+4)+C$

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

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