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

The integral from-1 to 0 of (2t)/((2+t^2)^3) is -5/72

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

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

- \frac{5}{72}

Decimal

- 0.06944 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{- 1}^{0} \frac{t}{( 2 + t ^{2} ) ^{3}} d t

Apply u-substitution

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

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

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

Apply the Power Rule

= 2 (- \frac{1}{2} [- \frac{1}{2 u ^{2}}]_{2}^{3})

Simplify

2 (- \frac{1}{2} [- \frac{1}{2 u ^{2}}]_{2}^{3}) : - [- \frac{1}{2 u ^{2}}]_{2}^{3}

= - [- \frac{1}{2 u ^{2}}]_{2}^{3}

Compute the boundaries:

\frac{5}{72}

= - \frac{5}{72}

What is the integral from-1 to 0 of (2t)/((2+t^2)^3) ?
The integral from-1 to 0 of (2t)/((2+t^2)^3) is -5/72