What is the integral from 0 to 5 of 8/(sqrt(7x+6)) ?

The integral from 0 to 5 of 8/(sqrt(7x+6)) is (8(2sqrt(41)-2sqrt(6)))/7

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integral from 0 to 5 of 8/(sqrt(7x+6))

\int_{0}^{5} \frac{8}{7 x + 6} d x

\frac{8 ( 2 41 - 2 6 )}{7}

Decimal

9.03687 \dots

Solution steps

\int_{0}^{5} \frac{8}{7 x + 6} d x

Take the constant out:

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

= 8 \cdot \int_{0}^{5} \frac{1}{7 x + 6} d x

Apply u-substitution

: \int_{6}^{41} \frac{1}{7 u} d u

= 8 \cdot \int_{6}^{41} \frac{1}{7 u} d u

Take the constant out:

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

= 8 \cdot \frac{1}{7} \cdot \int_{6}^{41} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{6}^{41}

= 8 \cdot \frac{1}{7} [2 u^{\frac{1}{2}}]_{6}^{41}

Simplify

8 \cdot \frac{1}{7} [2 u^{\frac{1}{2}}]_{6}^{41} : \frac{8}{7} [2 u]_{6}^{41}

= \frac{8}{7} [2 u]_{6}^{41}

Compute the boundaries:

241 - 26

= \frac{8}{7} (241 - 26)

Simplify

= \frac{8 ( 2 41 - 2 6 )}{7}

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What is the integral from 0 to 5 of 8/(sqrt(7x+6)) ?
The integral from 0 to 5 of 8/(sqrt(7x+6)) is (8(2sqrt(41)-2sqrt(6)))/7