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

The integral from 4 to 9 of x/(sqrt(9+x^2)) is 3sqrt(10)-5

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

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

310 - 5

Decimal Notation

4.48683 \dots

Solution steps

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

Apply u-substitution

: \int_{25}^{90} \frac{1}{2 u} d u

= \int_{25}^{90} \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_{25}^{90} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{25}^{90}

= \frac{1}{2} [2 u^{\frac{1}{2}}]_{25}^{90}

Apply radical rule:

= \frac{1}{2} [2 u]_{25}^{90}

Compute the boundaries:

610 - 10

= \frac{1}{2} (610 - 10)

Simplify

= 310 - 5

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What is the integral from 4 to 9 of x/(sqrt(9+x^2)) ?
The integral from 4 to 9 of x/(sqrt(9+x^2)) is 3sqrt(10)-5