Question

Integrate the following funtion :

Answer 1

\[\int\limits(3x^2 +1)/(x^2-1)^3 dx\]

Answer 2

partial fractions comes to my mind

Answer 3

by the way x^2-1 is factorable

Answer 4

x^2-1=(x-1)(x+1)

Answer 5

By partial fractions , it will be too long .

Answer 6

so you think there is a trick here?

Answer 7

well let's investigate

Answer 8

I hope so !

Answer 9

\[ 3x^2+1 = -(x^2-1) +4x^2\]

Answer 10

so the integrand becomes :\[\int\limits(-1/(x^2-1)^2dx + \int\limits4x^2/(x^2-1)^3dx\]

Answer 11

First integrand can be easily evaluated and the second one only partial fractions is striking me , I was hoping there could be a shorter method ...

Answer 12

\[3x^2+1=3(x^2-1)+4 \\ \int\limits_{}^{}\frac{3x^2+1}{(x^2-1)^3} dx=\int\limits_{}^{}\frac{3(x^2-1)+4}{(x^2-1)^3} dx=\int\limits \frac{3}{(x^2-1)^2} dx+\int\limits \frac{4}{(x^2-1)^3} dx\]
thought this doesn't seem to make the problem easier
still seems to need partial fractions or a trig sub