What is the integral of (x^3)/((x^4-1)^2) ?

The integral of (x^3)/((x^4-1)^2) is 1/(8(x^2+1))+1/(16(x+1))-1/(16(x-1))+C

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integral of (x^3)/((x^4-1)^2)

\int \frac{x ^{3}}{( x ^{4} - 1 ) ^{2}} d x

\frac{1}{8 ( x ^{2} + 1 )} + \frac{1}{16 ( x + 1 )} - \frac{1}{16 ( x - 1 )} + C

Solution steps

\int \frac{x ^{3}}{( x ^{4} - 1 ) ^{2}} d x

Take the partial fraction of

\frac{x ^{3}}{( x ^{4} - 1 ) ^{2}} : - \frac{x}{4 ( x ^{2} + 1 ) ^{2}} - \frac{1}{16 ( x + 1 ) ^{2}} + \frac{1}{16 ( x - 1 ) ^{2}}

= \int - \frac{x}{4 ( x ^{2} + 1 ) ^{2}} - \frac{1}{16 ( x + 1 ) ^{2}} + \frac{1}{16 ( x - 1 ) ^{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{x}{4 ( x ^{2} + 1 ) ^{2}} d x - \int \frac{1}{16 ( x + 1 ) ^{2}} d x + \int \frac{1}{16 ( x - 1 ) ^{2}} d x

\int \frac{x}{4 ( x ^{2} + 1 ) ^{2}} d x = - \frac{1}{8 ( x ^{2} + 1 )}

\int \frac{1}{16 ( x + 1 ) ^{2}} d x = - \frac{1}{16 ( x + 1 )}

\int \frac{1}{16 ( x - 1 ) ^{2}} d x = - \frac{1}{16 ( x - 1 )}

= - (- \frac{1}{8 ( x ^{2} + 1 )}) - (- \frac{1}{16 ( x + 1 )}) - \frac{1}{16 ( x - 1 )}

Simplify

= \frac{1}{8 ( x ^{2} + 1 )} + \frac{1}{16 ( x + 1 )} - \frac{1}{16 ( x - 1 )}

Add a constant to the solution

= \frac{1}{8 ( x ^{2} + 1 )} + \frac{1}{16 ( x + 1 )} - \frac{1}{16 ( x - 1 )} + C

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What is the integral of (x^3)/((x^4-1)^2) ?
The integral of (x^3)/((x^4-1)^2) is 1/(8(x^2+1))+1/(16(x+1))-1/(16(x-1))+C