What is the integral from 0 to 2 of x^2(1+8/(x^2+4)) ?

The integral from 0 to 2 of x^2(1+8/(x^2+4)) is -4pi+56/3

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integral from 0 to 2 of x^2(1+8/(x^2+4))

\int_{0}^{2} x^{2} (1 + \frac{8}{x ^{2} + 4}) d x

- 4 π + \frac{56}{3}

Decimal

6.10029 \dots

Solution steps

\int_{0}^{2} x^{2} (1 + \frac{8}{x ^{2} + 4}) d x

Expand

x^{2} (1 + \frac{8}{x ^{2} + 4}) : \frac{x ^{4} + 12 x ^{2}}{x ^{2} + 4}

= \int_{0}^{2} \frac{x ^{4} + 12 x ^{2}}{x ^{2} + 4} d x

Expand

\frac{x ^{4} + 12 x ^{2}}{x ^{2} + 4} : \frac{x ^{4}}{x ^{2} + 4} + \frac{12 x ^{2}}{x ^{2} + 4}

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} \frac{x ^{4}}{x ^{2} + 4} d x + \int_{0}^{2} \frac{12 x ^{2}}{x ^{2} + 4} d x

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

\int_{0}^{2} \frac{12 x ^{2}}{x ^{2} + 4} d x = 24 (- \frac{π}{4} + 1)

= 8 (- \frac{2}{3} + \frac{π}{4}) + 24 (- \frac{π}{4} + 1)

Simplify

8 (- \frac{2}{3} + \frac{π}{4}) + 24 (- \frac{π}{4} + 1) : - 4 π + \frac{56}{3}

= - 4 π + \frac{56}{3}

What is the integral from 0 to 2 of x^2(1+8/(x^2+4)) ?
The integral from 0 to 2 of x^2(1+8/(x^2+4)) is -4pi+56/3