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

The integral of (x^3)/(16-x^2) is -(x^2)/2-8ln|-x^2+16|+C

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

\int \frac{x ^{3}}{16 - x ^{2}} d x

- \frac{x ^{2}}{2} - 8 ln - x^{2} + 16 + C

Solution steps

\int \frac{x ^{3}}{16 - x ^{2}} d x

Long division

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

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

\int x d x = \frac{x ^{2}}{2}

\int \frac{16 x}{- x ^{2} + 16} d x = - 8 ln - x^{2} + 16

= - \frac{x ^{2}}{2} - 8 ln - x^{2} + 16

Add a constant to the solution

= - \frac{x ^{2}}{2} - 8 ln - x^{2} + 16 + C

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What is the integral of (x^3)/(16-x^2) ?
The integral of (x^3)/(16-x^2) is -(x^2)/2-8ln|-x^2+16|+C