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

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

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

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

- \frac{3}{20}

Solution steps

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

Apply u-substitution

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

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \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-2 to 1 of t/((t^2+1)^2) ?
The integral from-2 to 1 of t/((t^2+1)^2) is -3/20