A rational function integration Evaluate $$\int \frac{3x^2+1}{(x^2-1)^3}dx$$ I tried breaking the numerator in terms of the denominator but it didn't help much. I also tried a few substitutions buy most of them were useless. Please give some hints. Thanks.

Partial fractions will do the trick.
Try write $$\frac{(3x^2+1)}{(x^2-1)^3}=\frac{ax^2+bx+c}{(x-1)^3}+\frac{ex^2+fx+g}{(x+1)^3}.$$
Further note that $(x+1)^3-(x-1)^3$ leaves only the quadratic and constant terms as non-zero.
So $$\frac{1}{(x-1)^3}-\frac{1}{(x+1)^3}= \frac{-6x^2-2}{(x-1)^3(x+1)^3} =\frac {-1}{2} \cdot \frac{(3x^2+1)}{(x-1)^3(x+1)^3}.$$
The rest is trivial.
