What is the integral from 2 to 9 of 1/(sqrt(8x+9)) ?

The integral from 2 to 9 of 1/(sqrt(8x+9)) is 1

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integral from 2 to 9 of 1/(sqrt(8x+9))

\int_{2}^{9} \frac{1}{8 x + 9} d x

1

Solution steps

\int_{2}^{9} \frac{1}{8 x + 9} d x

Apply u-substitution

: \int_{25}^{81} \frac{1}{8 u} d u

= \int_{25}^{81} \frac{1}{8 u} 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_{25}^{81} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{25}^{81}

= \frac{1}{8} [2 u^{\frac{1}{2}}]_{25}^{81}

Apply radical rule:

= \frac{1}{8} [2 u]_{25}^{81}

Compute the boundaries:

8

= \frac{1}{8} \cdot 8

Simplify

= 1

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What is the integral from 2 to 9 of 1/(sqrt(8x+9)) ?
The integral from 2 to 9 of 1/(sqrt(8x+9)) is 1