What is the integral of (3x)/((x^2-1)^5) ?

The integral of (3x)/((x^2-1)^5) is

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integral of (3x)/((x^2-1)^5)

\int \frac{3 x}{( x ^{2} - 1 ) ^{5}} d x

3 (- \frac{5}{256 ( x + 1 )} - \frac{5}{256 ( x + 1 ) ^{2}} - \frac{1}{64 ( x + 1 ) ^{3}} - \frac{1}{128 ( x + 1 ) ^{4}} + \frac{5}{256 ( x - 1 )} - \frac{5}{256 ( x - 1 ) ^{2}} + \frac{1}{64 ( x - 1 ) ^{3}} - \frac{1}{128 ( x - 1 ) ^{4}}) + C

Solution steps

\int \frac{3 x}{( x ^{2} - 1 ) ^{5}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int \frac{x}{( x ^{2} - 1 ) ^{5}} d x

Take the partial fraction of

\frac{x}{( x ^{2} - 1 ) ^{5}} : \frac{5}{256 ( x + 1 ) ^{2}} + \frac{5}{128 ( x + 1 ) ^{3}} + \frac{3}{64 ( x + 1 ) ^{4}} + \frac{1}{32 ( x + 1 ) ^{5}} - \frac{5}{256 ( x - 1 ) ^{2}} + \frac{5}{128 ( x - 1 ) ^{3}} - \frac{3}{64 ( x - 1 ) ^{4}} + \frac{1}{32 ( x - 1 ) ^{5}}

= 3 \cdot \int \frac{5}{256 ( x + 1 ) ^{2}} + \frac{5}{128 ( x + 1 ) ^{3}} + \frac{3}{64 ( x + 1 ) ^{4}} + \frac{1}{32 ( x + 1 ) ^{5}} - \frac{5}{256 ( x - 1 ) ^{2}} + \frac{5}{128 ( x - 1 ) ^{3}} - \frac{3}{64 ( x - 1 ) ^{4}} + \frac{1}{32 ( x - 1 ) ^{5}} 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

= 3 (\int \frac{5}{256 ( x + 1 ) ^{2}} d x + \int \frac{5}{128 ( x + 1 ) ^{3}} d x + \int \frac{3}{64 ( x + 1 ) ^{4}} d x + \int \frac{1}{32 ( x + 1 ) ^{5}} d x - \int \frac{5}{256 ( x - 1 ) ^{2}} d x + \int \frac{5}{128 ( x - 1 ) ^{3}} d x - \int \frac{3}{64 ( x - 1 ) ^{4}} d x + \int \frac{1}{32 ( x - 1 ) ^{5}} d x)

\int \frac{5}{256 ( x + 1 ) ^{2}} d x = - \frac{5}{256 ( x + 1 )}

\int \frac{5}{128 ( x + 1 ) ^{3}} d x = - \frac{5}{256 ( x + 1 ) ^{2}}

\int \frac{3}{64 ( x + 1 ) ^{4}} d x = - \frac{1}{64 ( x + 1 ) ^{3}}

\int \frac{1}{32 ( x + 1 ) ^{5}} d x = - \frac{1}{128 ( x + 1 ) ^{4}}

\int \frac{5}{256 ( x - 1 ) ^{2}} d x = - \frac{5}{256 ( x - 1 )}

\int \frac{5}{128 ( x - 1 ) ^{3}} d x = - \frac{5}{256 ( x - 1 ) ^{2}}

\int \frac{3}{64 ( x - 1 ) ^{4}} d x = - \frac{1}{64 ( x - 1 ) ^{3}}

\int \frac{1}{32 ( x - 1 ) ^{5}} d x = - \frac{1}{128 ( x - 1 ) ^{4}}

= 3 (- \frac{5}{256 ( x + 1 )} - \frac{5}{256 ( x + 1 ) ^{2}} - \frac{1}{64 ( x + 1 ) ^{3}} - \frac{1}{128 ( x + 1 ) ^{4}} - (- \frac{5}{256 ( x - 1 )}) - \frac{5}{256 ( x - 1 ) ^{2}} - (- \frac{1}{64 ( x - 1 ) ^{3}}) - \frac{1}{128 ( x - 1 ) ^{4}})

Simplify

= 3 (- \frac{5}{256 ( x + 1 )} - \frac{5}{256 ( x + 1 ) ^{2}} - \frac{1}{64 ( x + 1 ) ^{3}} - \frac{1}{128 ( x + 1 ) ^{4}} + \frac{5}{256 ( x - 1 )} - \frac{5}{256 ( x - 1 ) ^{2}} + \frac{1}{64 ( x - 1 ) ^{3}} - \frac{1}{128 ( x - 1 ) ^{4}})

Add a constant to the solution

= 3 (- \frac{5}{256 ( x + 1 )} - \frac{5}{256 ( x + 1 ) ^{2}} - \frac{1}{64 ( x + 1 ) ^{3}} - \frac{1}{128 ( x + 1 ) ^{4}} + \frac{5}{256 ( x - 1 )} - \frac{5}{256 ( x - 1 ) ^{2}} + \frac{1}{64 ( x - 1 ) ^{3}} - \frac{1}{128 ( x - 1 ) ^{4}}) + C

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