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

The integral from 1 to 3 of x/((x^2+9)^3) is 7/4050

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integral from 1 to 3 of x/((x^2+9)^3)

\int_{1}^{3} \frac{x}{( x ^{2} + 9 ) ^{3}} d x

\frac{7}{4050}

Decimal

0.00172 \dots

Solution steps

\int_{1}^{3} \frac{x}{( x ^{2} + 9 ) ^{3}} d x

Apply u-substitution

: \int_{10}^{18} \frac{1}{2 u ^{3}} d u

= \int_{10}^{18} \frac{1}{2 u ^{3}} 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_{10}^{18} \frac{1}{u ^{3}} d u

Apply the Power Rule

: [- \frac{1}{2 u ^{2}}]_{10}^{18}

= \frac{1}{2} [- \frac{1}{2 u ^{2}}]_{10}^{18}

Compute the boundaries:

\frac{7}{2025}

= \frac{1}{2} \cdot \frac{7}{2025}

Simplify

= \frac{7}{4050}

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What is the integral from 1 to 3 of x/((x^2+9)^3) ?
The integral from 1 to 3 of x/((x^2+9)^3) is 7/4050