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

The integral from 0 to 2 of x/((1+x^2)^3) is 6/25

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

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

\frac{6}{25}

Decimal

0.24

Solution steps

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

Apply u-substitution

: \int_{1}^{5} \frac{1}{2 u ^{3}} d u

= \int_{1}^{5} \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

= \frac{1}{2} \cdot \int_{1}^{5} \frac{1}{u ^{3}} d u

Apply the Power Rule

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

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

Compute the boundaries:

\frac{12}{25}

= \frac{1}{2} \cdot \frac{12}{25}

Simplify

= \frac{6}{25}

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What is the integral from 0 to 2 of x/((1+x^2)^3) ?
The integral from 0 to 2 of x/((1+x^2)^3) is 6/25