What is the integral of (x^3)/((x^2+16)^3) ?

The integral of (x^3)/((x^2+16)^3) is -1/(2(x^2+16))+4/((x^2+16)^2)+C

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integral of (x^3)/((x^2+16)^3)

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

- \frac{1}{2 ( x ^{2} + 16 )} + \frac{4}{( x ^{2} + 16 ) ^{2}} + C

Solution steps

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

Take the partial fraction of

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

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

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

\int \frac{16 x}{( x ^{2} + 16 ) ^{3}} d x = - \frac{4}{( x ^{2} + 16 ) ^{2}}

= - \frac{1}{2 ( x ^{2} + 16 )} - (- \frac{4}{( x ^{2} + 16 ) ^{2}})

Simplify

= - \frac{1}{2 ( x ^{2} + 16 )} + \frac{4}{( x ^{2} + 16 ) ^{2}}

Add a constant to the solution

= - \frac{1}{2 ( x ^{2} + 16 )} + \frac{4}{( x ^{2} + 16 ) ^{2}} + C

What is the integral of (x^3)/((x^2+16)^3) ?
The integral of (x^3)/((x^2+16)^3) is -1/(2(x^2+16))+4/((x^2+16)^2)+C