What is the integral from 0 to 1 of (2x)/((x^2+1)^3) ?

The integral from 0 to 1 of (2x)/((x^2+1)^3) is 3/8

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integral from 0 to 1 of (2x)/((x^2+1)^3)

\int_{0}^{1} \frac{2 x}{( x ^{2} + 1 ) ^{3}} d x

\frac{3}{8}

Decimal Notation

0.375

Solution steps

\int_{0}^{1} \frac{2 x}{( x ^{2} + 1 ) ^{3}} 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 ) ^{3}} d x

Apply u-substitution

: \int_{1}^{2} \frac{1}{2 u ^{3}} d u

= 2 \cdot \int_{1}^{2} \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

= 2 \cdot \frac{1}{2} \cdot \int_{1}^{2} \frac{1}{u ^{3}} d u

Apply the Power Rule

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

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

Simplify

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

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

Compute the boundaries:

\frac{3}{8}

= \frac{3}{8}

\int (\frac{2}{1 + 2 x}) d x

\frac{d}{d y} (\frac{3 x}{2 y})

\int \frac{1}{sin ^{2} ( x ) + tan ^{2} ( x )} d x

y = - x, y = 2 - x, y = 0

y = \frac{3 x ^{2}}{2 x + 4}

What is the integral from 0 to 1 of (2x)/((x^2+1)^3) ?
The integral from 0 to 1 of (2x)/((x^2+1)^3) is 3/8