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

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

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integral from 0 to 1 of (2x)/((x^2+1)^2)

\int_{0}^{1} \frac{2 x}{( x ^{2} + 1 ) ^{2}} d x

\frac{1}{2}

Decimal

0.5

Solution steps

\int_{0}^{1} \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_{0}^{1} \frac{x}{( x ^{2} + 1 ) ^{2}} d x

Apply u-substitution

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

Apply the Power Rule

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

Simplify

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

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

Compute the boundaries:

\frac{1}{2}

= \frac{1}{2}

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