What is the integral of 8/(x(x-8)^2) ?

The integral of 8/(x(x-8)^2) is 1/8 ln|x|-1/8 ln|x-8|-1/(x-8)+C

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integral of 8/(x(x-8)^2)

\int \frac{8}{x ( x - 8 ) ^{2}} d x

\frac{1}{8} ln ∣ x ∣ - \frac{1}{8} ln ∣ x - 8 ∣ - \frac{1}{x - 8} + C

Solution steps

\int \frac{8}{x ( x - 8 ) ^{2}} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{x ( x - 8 ) ^{2}} d x

Take the partial fraction of

\frac{1}{x ( x - 8 ) ^{2}} : \frac{1}{64 x} - \frac{1}{64 ( x - 8 )} + \frac{1}{8 ( x - 8 ) ^{2}}

= 8 \cdot \int \frac{1}{64 x} - \frac{1}{64 ( x - 8 )} + \frac{1}{8 ( x - 8 ) ^{2}} d x

Apply the Sum Rule:

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

= 8 (\int \frac{1}{64 x} d x - \int \frac{1}{64 ( x - 8 )} d x + \int \frac{1}{8 ( x - 8 ) ^{2}} d x)

\int \frac{1}{64 x} d x = \frac{1}{64} ln ∣ x ∣

\int \frac{1}{64 ( x - 8 )} d x = \frac{1}{64} ln ∣ x - 8 ∣

\int \frac{1}{8 ( x - 8 ) ^{2}} d x = - \frac{1}{8 ( x - 8 )}

= 8 (\frac{1}{64} ln ∣ x ∣ - \frac{1}{64} ln ∣ x - 8 ∣ - \frac{1}{8 ( x - 8 )})

Simplify

8 (\frac{1}{64} ln ∣ x ∣ - \frac{1}{64} ln ∣ x - 8 ∣ - \frac{1}{8 ( x - 8 )}) : \frac{1}{8} ln ∣ x ∣ - \frac{1}{8} ln ∣ x - 8 ∣ - \frac{1}{x - 8}

= \frac{1}{8} ln ∣ x ∣ - \frac{1}{8} ln ∣ x - 8 ∣ - \frac{1}{x - 8}

Add a constant to the solution

= \frac{1}{8} ln ∣ x ∣ - \frac{1}{8} ln ∣ x - 8 ∣ - \frac{1}{x - 8} + C

\int_{1}^{\infty} \frac{1}{x ^{1.01}} d x

\frac{d}{d x} (\frac{1}{x + 10})

\frac{d}{d x} (12 ln (x))

\frac{\partial}{\partial y} (\frac{x}{x - y})

y^{^{'}} = y sin (x)

What is the integral of 8/(x(x-8)^2) ?
The integral of 8/(x(x-8)^2) is 1/8 ln|x|-1/8 ln|x-8|-1/(x-8)+C