What is the integral from 0 to 6 of x/(sqrt(25+4x)) ?

The integral from 0 to 6 of x/(sqrt(25+4x)) is 17/6

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integral from 0 to 6 of x/(sqrt(25+4x))

\int_{0}^{6} \frac{x}{25 + 4 x} d x

\frac{17}{6}

Decimal

2.83333 \dots

Solution steps

\int_{0}^{6} \frac{x}{25 + 4 x} d x

Apply u-substitution

: \int_{5}^{7} \frac{u ^{2} - 25}{8} d u

= \int_{5}^{7} \frac{u ^{2} - 25}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int_{5}^{7} u^{2} - 25 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{1}{8} (\int_{5}^{7} u^{2} d u - \int_{5}^{7} 25 d u)

\int_{5}^{7} u^{2} d u = \frac{218}{3}

\int_{5}^{7} 25 d u = 50

= \frac{1}{8} (\frac{218}{3} - 50)

Simplify

\frac{1}{8} (\frac{218}{3} - 50) : \frac{17}{6}

= \frac{17}{6}

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What is the integral from 0 to 6 of x/(sqrt(25+4x)) ?
The integral from 0 to 6 of x/(sqrt(25+4x)) is 17/6