What is the integral from-1 to 0 of (3x)/((2+x^2)^4) ?

The integral from-1 to 0 of (3x)/((2+x^2)^4) is -19/432

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integral from-1 to 0 of (3x)/((2+x^2)^4)

\int_{- 1}^{0} \frac{3 x}{( 2 + x ^{2} ) ^{4}} d x

- \frac{19}{432}

Decimal

- 0.04398 \dots

Solution steps

\int_{- 1}^{0} \frac{3 x}{( 2 + x ^{2} ) ^{4}} d x

Take the constant out:

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

= 3 \cdot \int_{- 1}^{0} \frac{x}{( 2 + x ^{2} ) ^{4}} d x

Apply u-substitution

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

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

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

= 3 (- \int_{2}^{3} \frac{1}{2 u ^{4}} d u)

Take the constant out:

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

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

Apply the Power Rule

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

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

Simplify

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

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

Compute the boundaries:

\frac{19}{648}

= - \frac{3}{2} \cdot \frac{19}{648}

Simplify

= - \frac{19}{432}

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What is the integral from-1 to 0 of (3x)/((2+x^2)^4) ?
The integral from-1 to 0 of (3x)/((2+x^2)^4) is -19/432