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

The integral of (3x^3-x)/((x^2+1)^5) is -1/(2(x^2+1)^3)+1/(2(x^2+1)^4)+C

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

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

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

Solution steps

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

Expand

\frac{3 x ^{3} - x}{( x ^{2} + 1 ) ^{5}} : \frac{3 x ^{3}}{( x ^{2} + 1 ) ^{5}} - \frac{x}{( x ^{2} + 1 ) ^{5}}

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

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

\int \frac{3 x ^{3}}{( x ^{2} + 1 ) ^{5}} d x = 3 (- \frac{1}{6 ( x ^{2} + 1 ) ^{3}} + \frac{1}{8 ( x ^{2} + 1 ) ^{4}})

\int \frac{x}{( x ^{2} + 1 ) ^{5}} d x = - \frac{1}{8 ( x ^{2} + 1 ) ^{4}}

= 3 (- \frac{1}{6 ( x ^{2} + 1 ) ^{3}} + \frac{1}{8 ( x ^{2} + 1 ) ^{4}}) - (- \frac{1}{8 ( x ^{2} + 1 ) ^{4}})

Simplify

3 (- \frac{1}{6 ( x ^{2} + 1 ) ^{3}} + \frac{1}{8 ( x ^{2} + 1 ) ^{4}}) - (- \frac{1}{8 ( x ^{2} + 1 ) ^{4}}) : - \frac{1}{2 ( x ^{2} + 1 ) ^{3}} + \frac{1}{2 ( x ^{2} + 1 ) ^{4}}

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

Add a constant to the solution

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

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What is the integral of (3x^3-x)/((x^2+1)^5) ?
The integral of (3x^3-x)/((x^2+1)^5) is -1/(2(x^2+1)^3)+1/(2(x^2+1)^4)+C