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

The integral from-1 to 0 of (2t)/((3+t^2)^3) is -7/288

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

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

- \frac{7}{288}

Decimal Notation

- 0.02430 \dots

Solution steps

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

Apply u-substitution

: \int_{4}^{3} \frac{1}{2 u ^{3}} d u

= 2 \cdot \int_{4}^{3} \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_{3}^{4} \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_{3}^{4} \frac{1}{u ^{3}} d u)

Apply the Power Rule

: [- \frac{1}{2 u ^{2}}]_{3}^{4}

= 2 (- \frac{1}{2} [- \frac{1}{2 u ^{2}}]_{3}^{4})

Simplify

2 (- \frac{1}{2} [- \frac{1}{2 u ^{2}}]_{3}^{4}) : - [- \frac{1}{2 u ^{2}}]_{3}^{4}

= - [- \frac{1}{2 u ^{2}}]_{3}^{4}

Compute the boundaries:

\frac{7}{288}

= - \frac{7}{288}

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What is the integral from-1 to 0 of (2t)/((3+t^2)^3) ?
The integral from-1 to 0 of (2t)/((3+t^2)^3) is -7/288