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

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Answer 1 · 2017-01-20T13:15:23

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

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

Substitute $x^2=t," so that, "2xdx=dt, or, xdx=1/2dt$

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

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

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

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

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

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

Enjoy Maths.!

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