# How to integ 2x^2 / (1-x^2) using partial fraction?

how to integ 2x^2 / (1-x^2)
using partial fracton??

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ah....i forgot to type the answer=-2x + ln(1+x) - ln(1-x)

HallsofIvy
First divide to get a "mixed" fraction: $\frac{2x^}{1- x^2}= -2- \frac{2}{1-x^2}$. Now use partial fractions on the second term.
$$\frac{2x^{2}}{1-x^{2}} = \frac{2x^{2}-2+2}{1-x^{2}} = \frac{-2(1-x^{2})}{1-x^{2}} +\frac{2}{1-x^{2}} = -2 +\frac{2}{1-x^{2}}$$