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

The integral of (x^3)/((1+2x^2)^2) is 1/8 ln|2x^2+1|+1/(8(2x^2+1))+C

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

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What is the integral of (x^3)/((1+2x^2)^2) ?
The integral of (x^3)/((1+2x^2)^2) is 1/8 ln|2x^2+1|+1/(8(2x^2+1))+C