What is the integral from 0 to 12 of x/(sqrt(4+5x)) ?

The integral from 0 to 12 of x/(sqrt(4+5x)) is 288/25

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integral from 0 to 12 of x/(sqrt(4+5x))

\int_{0}^{12} \frac{x}{4 + 5 x} d x

\frac{288}{25}

Decimal

11.52

Solution steps

\int_{0}^{12} \frac{x}{4 + 5 x} d x

Apply u-substitution

: \int_{2}^{8} \frac{2 ( u ^{2} - 4 )}{25} d u

= \int_{2}^{8} \frac{2 ( u ^{2} - 4 )}{25} d u

Take the constant out:

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

= \frac{2}{25} \cdot \int_{2}^{8} 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

= \frac{2}{25} (\int_{2}^{8} u^{2} d u - \int_{2}^{8} 4 d u)

\int_{2}^{8} u^{2} d u = 168

\int_{2}^{8} 4 d u = 24

= \frac{2}{25} (168 - 24)

Simplify

\frac{2}{25} (168 - 24) : \frac{288}{25}

= \frac{288}{25}

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What is the integral from 0 to 12 of x/(sqrt(4+5x)) ?
The integral from 0 to 12 of x/(sqrt(4+5x)) is 288/25