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

The integral from 0 to 2 of (2x)/((x^2+4)^2) is 1/8

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

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

\frac{1}{8}

Decimal

0.125

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{2} \frac{x}{( x ^{2} + 4 ) ^{2}} d x

Apply u-substitution

: \int_{4}^{8} \frac{1}{2 u ^{2}} d u

= 2 \cdot \int_{4}^{8} \frac{1}{2 u ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \frac{1}{2} \cdot \int_{4}^{8} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{4}^{8}

= 2 \cdot \frac{1}{2} [- \frac{1}{u}]_{4}^{8}

Simplify

2 \cdot \frac{1}{2} [- \frac{1}{u}]_{4}^{8} : [- \frac{1}{u}]_{4}^{8}

= [- \frac{1}{u}]_{4}^{8}

Compute the boundaries:

\frac{1}{8}

= \frac{1}{8}

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What is the integral from 0 to 2 of (2x)/((x^2+4)^2) ?
The integral from 0 to 2 of (2x)/((x^2+4)^2) is 1/8