What is the integral from 0 to 3 of x/(sqrt(x^2+9)) ?

The integral from 0 to 3 of x/(sqrt(x^2+9)) is 3(sqrt(2)-1)

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integral from 0 to 3 of x/(sqrt(x^2+9))

\int_{0}^{3} \frac{x}{x ^{2} + 9} d x

3 (2 - 1)

Solution steps

\int_{0}^{3} \frac{x}{x ^{2} + 9} d x

Apply u-substitution

: \int_{9}^{18} \frac{1}{2 u} d u

= \int_{9}^{18} \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_{9}^{18} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{9}^{18}

= \frac{1}{2} [2 u^{\frac{1}{2}}]_{9}^{18}

Apply radical rule:

= \frac{1}{2} [2 u]_{9}^{18}

Compute the boundaries:

62 - 6

= \frac{1}{2} (62 - 6)

Simplify

= 3 (2 - 1)

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What is the integral from 0 to 3 of x/(sqrt(x^2+9)) ?
The integral from 0 to 3 of x/(sqrt(x^2+9)) is 3(sqrt(2)-1)