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

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

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

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

2 ln (2) + π

Decimal

4.52788 \dots

Solution steps

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

Expand

\frac{4 x + 8}{x ^{2} + 4} : \frac{4 x}{x ^{2} + 4} + \frac{8}{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{4 x}{x ^{2} + 4} d x + \int_{0}^{2} \frac{8}{x ^{2} + 4} d x

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

\int_{0}^{2} \frac{8}{x ^{2} + 4} d x = π

= 2 ln (2) + π

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