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

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

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

integral

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

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

Solution steps

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

Take the partial fraction of

\frac{x ^{3}}{( x ^{2} - 1 ) ^{3}} : - \frac{3}{16 ( x + 1 ) ^{2}} + \frac{1}{8 ( x + 1 ) ^{3}} + \frac{3}{16 ( x - 1 ) ^{2}} + \frac{1}{8 ( x - 1 ) ^{3}}

= \int - \frac{3}{16 ( x + 1 ) ^{2}} + \frac{1}{8 ( x + 1 ) ^{3}} + \frac{3}{16 ( x - 1 ) ^{2}} + \frac{1}{8 ( x - 1 ) ^{3}} 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{3}{16 ( x + 1 ) ^{2}} d x + \int \frac{1}{8 ( x + 1 ) ^{3}} d x + \int \frac{3}{16 ( x - 1 ) ^{2}} d x + \int \frac{1}{8 ( x - 1 ) ^{3}} d x

\int \frac{3}{16 ( x + 1 ) ^{2}} d x = - \frac{3}{16 ( x + 1 )}

\int \frac{1}{8 ( x + 1 ) ^{3}} d x = - \frac{1}{16 ( x + 1 ) ^{2}}

\int \frac{3}{16 ( x - 1 ) ^{2}} d x = - \frac{3}{16 ( x - 1 )}

\int \frac{1}{8 ( x - 1 ) ^{3}} d x = - \frac{1}{16 ( x - 1 ) ^{2}}

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

Simplify

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

Add a constant to the solution

= \frac{3}{16 ( x + 1 )} - \frac{1}{16 ( x + 1 ) ^{2}} - \frac{3}{16 ( x - 1 )} - \frac{1}{16 ( x - 1 ) ^{2}} + C

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