What is the integral from 2 to 3 of x/(sqrt(9x^2-9)) ?

The integral from 2 to 3 of x/(sqrt(9x^2-9)) is (2sqrt(2)-sqrt(3))/3

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integral from 2 to 3 of x/(sqrt(9x^2-9))

\int_{2}^{3} \frac{x}{9 x ^{2} - 9} d x

\frac{2 2 - 3}{3}

Solution steps

\int_{2}^{3} \frac{x}{9 x ^{2} - 9} d x

Apply u-substitution

: \int_{4}^{9} \frac{1}{6 u - 1} d u

= \int_{4}^{9} \frac{1}{6 u - 1} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int_{4}^{9} \frac{1}{u - 1} d u

Apply u-substitution

: \int_{3}^{22} 2 d v

= \frac{1}{6} \cdot \int_{3}^{22} 2 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{6} [2 v]_{3}^{22}

Compute the boundaries:

42 - 23

= \frac{1}{6} (42 - 23)

Simplify

= \frac{2 2 - 3}{3}

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What is the integral from 2 to 3 of x/(sqrt(9x^2-9)) ?
The integral from 2 to 3 of x/(sqrt(9x^2-9)) is (2sqrt(2)-sqrt(3))/3