What is the integral from 0 to 2 of (x^3)/(16-x^2) ?

The integral from 0 to 2 of (x^3)/(16-x^2) is -2+8ln(4/3)

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integral from 0 to 2 of (x^3)/(16-x^2)

\int_{0}^{2} \frac{x ^{3}}{16 - x ^{2}} d x

- 2 + 8 ln (\frac{4}{3})

Decimal

0.30145 \dots

Solution steps

\int_{0}^{2} \frac{x ^{3}}{16 - x ^{2}} d x

Long division

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

= \int_{0}^{2} - 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_{0}^{2} x d x + \int_{0}^{2} \frac{16 x}{- x ^{2} + 16} d x

\int_{0}^{2} x d x = 2

\int_{0}^{2} \frac{16 x}{- x ^{2} + 16} d x = 8 ln (\frac{4}{3})

= - 2 + 8 ln (\frac{4}{3})

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\int_{0}^{1} (arctan (x)) d x

What is the integral from 0 to 2 of (x^3)/(16-x^2) ?
The integral from 0 to 2 of (x^3)/(16-x^2) is -2+8ln(4/3)