What is the integral from 0 to 12 of x/(sqrt(x+4)) ?

The integral from 0 to 12 of x/(sqrt(x+4)) is 64/3

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integral from 0 to 12 of x/(sqrt(x+4))

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

\frac{64}{3}

Decimal

21.33333 \dots

Solution steps

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

Apply u-substitution

: \int_{2}^{4} 2 (u^{2} - 4) d u

= \int_{2}^{4} 2 (u^{2} - 4) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{2}^{4} u^{2} - 4 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int_{2}^{4} u^{2} d u - \int_{2}^{4} 4 d u)

\int_{2}^{4} u^{2} d u = \frac{56}{3}

\int_{2}^{4} 4 d u = 8

= 2 (\frac{56}{3} - 8)

Simplify

2 (\frac{56}{3} - 8) : \frac{64}{3}

= \frac{64}{3}

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What is the integral from 0 to 12 of x/(sqrt(x+4)) ?
The integral from 0 to 12 of x/(sqrt(x+4)) is 64/3