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

The integral from 2 to 3 of (2x)/((x^2+1)^2) is 1/10

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

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

\frac{1}{10}

Decimal

0.1

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

= 2 \cdot \int_{5}^{10} \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}^{10} \frac{1}{u ^{2}} d u

Apply the Power Rule

= 2 \cdot \frac{1}{2} [- \frac{1}{u}]_{5}^{10}

Simplify

2 \cdot \frac{1}{2} [- \frac{1}{u}]_{5}^{10} : [- \frac{1}{u}]_{5}^{10}

= [- \frac{1}{u}]_{5}^{10}

Compute the boundaries:

\frac{1}{10}

= \frac{1}{10}

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What is the integral from 2 to 3 of (2x)/((x^2+1)^2) ?
The integral from 2 to 3 of (2x)/((x^2+1)^2) is 1/10