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

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

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

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

Take the partial fraction of

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

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