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

The integral from-1 to 2 of x/((x^2+1)^2) is 3/20

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

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

\frac{3}{20}

Decimal Notation

0.15

Solution steps

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

Apply u-substitution

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

= \int_{2}^{5} \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

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{3}{10}

= \frac{1}{2} \cdot \frac{3}{10}

Simplify

= \frac{3}{20}

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What is the integral from-1 to 2 of x/((x^2+1)^2) ?
The integral from-1 to 2 of x/((x^2+1)^2) is 3/20