What is the integral from 0 to 4 of x/(sqrt(x^2+9)) ?

The integral from 0 to 4 of x/(sqrt(x^2+9)) is 2

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integral from 0 to 4 of x/(sqrt(x^2+9))

\int_{0}^{4} \frac{x}{x ^{2} + 9} d x

2

Solution steps

\int_{0}^{4} \frac{x}{x ^{2} + 9} d x

Apply u-substitution

: \int_{9}^{25} \frac{1}{2 u} d u

= \int_{9}^{25} \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}^{25} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{9}^{25}

= \frac{1}{2} [2 u^{\frac{1}{2}}]_{9}^{25}

Apply radical rule:

= \frac{1}{2} [2 u]_{9}^{25}

Compute the boundaries:

4

= \frac{1}{2} \cdot 4

Simplify

= 2

\int_{2}^{\infty} \frac{2}{x ^{2} - 1} d x

x \to + 0 lim (ln (x^{2}))

\frac{2 s + 1}{2 s ^{2} + s + 1}

\int - 2 x + 12 x^{2} - 4 x^{3} + 14 d x

\int \frac{2 sin ^{3} ( x )}{x} d x

What is the integral from 0 to 4 of x/(sqrt(x^2+9)) ?
The integral from 0 to 4 of x/(sqrt(x^2+9)) is 2