What is the integral from 2 to 4 of x/((x^2+6)^3) ?

The integral from 2 to 4 of x/((x^2+6)^3) is 6/3025

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integral from 2 to 4 of x/((x^2+6)^3)

\int_{2}^{4} \frac{x}{( x ^{2} + 6 ) ^{3}} d x

\frac{6}{3025}

Decimal

0.00198 \dots

Solution steps

\int_{2}^{4} \frac{x}{( x ^{2} + 6 ) ^{3}} d x

Apply u-substitution

= \int_{10}^{22} \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}^{22} \frac{1}{u ^{3}} d u

Apply the Power Rule

= \frac{1}{2} [- \frac{1}{2 u ^{2}}]_{10}^{22}

Compute the boundaries:

\frac{12}{3025}

= \frac{1}{2} \cdot \frac{12}{3025}

Simplify

= \frac{6}{3025}

What is the integral from 2 to 4 of x/((x^2+6)^3) ?
The integral from 2 to 4 of x/((x^2+6)^3) is 6/3025