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.#

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