What is the integral from 3 to 5 of x/(sqrt(x^2-9)) ?

The integral from 3 to 5 of x/(sqrt(x^2-9)) is 4

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integral from 3 to 5 of x/(sqrt(x^2-9))

\int_{3}^{5} \frac{x}{x ^{2} - 9} d x

4

Solution steps

\int_{3}^{5} \frac{x}{x ^{2} - 9} d x

Apply u-substitution

: \int_{0}^{16} \frac{1}{2 u} d u

= \int_{0}^{16} \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{16} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{0}^{16}

= \frac{1}{2} [2 u^{\frac{1}{2}}]_{0}^{16}

Apply radical rule:

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

= \frac{1}{2} [2 u]_{0}^{16}

Compute the boundaries:

8

= \frac{1}{2} \cdot 8

Simplify

= 4

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What is the integral from 3 to 5 of x/(sqrt(x^2-9)) ?
The integral from 3 to 5 of x/(sqrt(x^2-9)) is 4