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

The integral from 0 to 1 of x/((x^2-4)^3) is -7/576

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

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

- \frac{7}{576}

Decimal

- 0.01215 \dots

Solution steps

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

Apply u-substitution

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

= \int_{- 4}^{- 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

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

Apply the Power Rule

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

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

Compute the boundaries:

- \frac{7}{288}

= \frac{1}{2} (- \frac{7}{288})

Simplify

= - \frac{7}{576}

\int_{5000}^{10000} 48 - 0.12 x d x

\int_{- 7}^{28} x 3 1 - x d x

\int_{1}^{2} - \frac{1}{z} d z

\int_{0}^{1} 4 x^{2} - 4 x^{4} d x

\int_{0}^{1} x^{3} - 4 x d x

What is the integral from 0 to 1 of x/((x^2-4)^3) ?
The integral from 0 to 1 of x/((x^2-4)^3) is -7/576