How do you integrate $\frac{3 x}{x^{2} \cdot (x^{2} + 1)}$ $(3x) / (x^2 * (x^2+1) )$ using partial fractions?

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Answer 1 · 2017-01-20T13:15:23

$\frac{3}{2} ln {\frac{x^{2}}{x^{2} + 1}} + C .$ $3/2ln{x^2/(x^2+1)}+C.$

Let $I = \int \frac{3 x}{x^{2} (x^{2} + 1)} d x$ $I=int(3x)/{x^2(x^2+1)}dx$

Substitute $x^{2} = t, so that, 2 x d x = d t, or, x d x = \frac{1}{2} d t$ $x^2=t," so that, "2xdx=dt, or, xdx=1/2dt$

$:. I=3int(1/2)/{t(t+1)}dt=3/2int{(t+1)-t}/{t(t+1)}dt$

$=3/2int[(t+1)/{t(t+1)}-t/{t(t+1)}]dt$

$=3/2int[1/t-1/(t+1)]dt$

$=3/2[ln|t|-ln|t+1|]$

$=3/2ln|t/(t+1)|, and, because, t=x^2,$

$I=3/2ln{x^2/(x^2+1)}+C.$

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How do you integrate (3x) / (x^2 * (x^2+1) )3xx2⋅(x2+1)(3x) / (x^2 * (x^2+1) ) using partial fractions?