What is the integral from 0 to 8 of x/((x+8)^3) ?

The integral from 0 to 8 of x/((x+8)^3) is 1/64

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integral from 0 to 8 of x/((x+8)^3)

\int_{0}^{8} \frac{x}{( x + 8 ) ^{3}} d x

\frac{1}{64}

Decimal Notation

0.015625

Solution steps

\int_{0}^{8} \frac{x}{( x + 8 ) ^{3}} d x

Apply u-substitution

: \int_{8}^{16} \frac{u - 8}{u ^{3}} d u

= \int_{8}^{16} \frac{u - 8}{u ^{3}} d u

Expand

\frac{u - 8}{u ^{3}} : \frac{1}{u ^{2}} - \frac{8}{u ^{3}}

= \int_{8}^{16} \frac{1}{u ^{2}} - \frac{8}{u ^{3}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{8}^{16} \frac{1}{u ^{2}} d u - \int_{8}^{16} \frac{8}{u ^{3}} d u

\int_{8}^{16} \frac{1}{u ^{2}} d u = \frac{1}{16}

\int_{8}^{16} \frac{8}{u ^{3}} d u = \frac{3}{64}

= \frac{1}{16} - \frac{3}{64}

Simplify

\frac{1}{16} - \frac{3}{64} : \frac{1}{64}

= \frac{1}{64}

\frac{d}{d x} ((3 x^{2} - 5 x) e^{x})

\int \frac{x}{x - 1} d x

y = \frac{e ^{x}}{5 x ^{2}}

\frac{d}{d x} ((2 x + 3)^{1.4})

\frac{\partial}{\partial x} (x + y^{\frac{1}{2}})

What is the integral from 0 to 8 of x/((x+8)^3) ?
The integral from 0 to 8 of x/((x+8)^3) is 1/64