What is the integral from 3 to 8 of 1/(sqrt(9x+9)) ?

The integral from 3 to 8 of 1/(sqrt(9x+9)) is 2/3

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integral from 3 to 8 of 1/(sqrt(9x+9))

\int_{3}^{8} \frac{1}{9 x + 9} d x

\frac{2}{3}

Decimal

0.66666 \dots

Solution steps

\int_{3}^{8} \frac{1}{9 x + 9} d x

Apply u-substitution

: \int_{36}^{81} \frac{1}{9 u} d u

= \int_{36}^{81} \frac{1}{9 u} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int_{36}^{81} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{36}^{81}

= \frac{1}{9} [2 u^{\frac{1}{2}}]_{36}^{81}

Apply radical rule:

a^{\frac{1}{n}} = n a

= \frac{1}{9} [2 u]_{36}^{81}

Compute the boundaries:

6

= \frac{1}{9} \cdot 6

Simplify

= \frac{2}{3}

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What is the integral from 3 to 8 of 1/(sqrt(9x+9)) ?
The integral from 3 to 8 of 1/(sqrt(9x+9)) is 2/3