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

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

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

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

- \frac{5}{72}

Decimal

- 0.06944 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Simplify

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

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

Compute the boundaries:

- \frac{5}{72}

= - \frac{5}{72}

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