What is the integral from 1 to 3 of (2x)/((x^2+4)^2) ?

The integral from 1 to 3 of (2x)/((x^2+4)^2) is 8/65

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integral from 1 to 3 of (2x)/((x^2+4)^2)

\int_{1}^{3} \frac{2 x}{( x ^{2} + 4 ) ^{2}} d x

\frac{8}{65}

Solution steps

\int_{1}^{3} \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_{1}^{3} \frac{x}{( x ^{2} + 4 ) ^{2}} d x

Apply u-substitution

: \int_{5}^{13} \frac{1}{2 u ^{2}} d u

= 2 \cdot \int_{5}^{13} \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_{5}^{13} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{5}^{13}

= 2 \cdot \frac{1}{2} [- \frac{1}{u}]_{5}^{13}

Simplify

2 \cdot \frac{1}{2} [- \frac{1}{u}]_{5}^{13} : [- \frac{1}{u}]_{5}^{13}

= [- \frac{1}{u}]_{5}^{13}

Compute the boundaries:

\frac{8}{65}

= \frac{8}{65}

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What is the integral from 1 to 3 of (2x)/((x^2+4)^2) ?
The integral from 1 to 3 of (2x)/((x^2+4)^2) is 8/65