What is the integral of 6/(x(x-6)^2) ?

The integral of 6/(x(x-6)^2) is 1/6 ln|x|-1/6 ln|x-6|-1/(x-6)+C

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integral of 6/(x(x-6)^2)

\int \frac{6}{x ( x - 6 ) ^{2}} d x

\frac{1}{6} ln ∣ x ∣ - \frac{1}{6} ln ∣ x - 6 ∣ - \frac{1}{x - 6} + C

Solution steps

\int \frac{6}{x ( x - 6 ) ^{2}} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{x ( x - 6 ) ^{2}} d x

Take the partial fraction of

\frac{1}{x ( x - 6 ) ^{2}} : \frac{1}{36 x} - \frac{1}{36 ( x - 6 )} + \frac{1}{6 ( x - 6 ) ^{2}}

= 6 \cdot \int \frac{1}{36 x} - \frac{1}{36 ( x - 6 )} + \frac{1}{6 ( x - 6 ) ^{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

= 6 (\int \frac{1}{36 x} d x - \int \frac{1}{36 ( x - 6 )} d x + \int \frac{1}{6 ( x - 6 ) ^{2}} d x)

\int \frac{1}{36 x} d x = \frac{1}{36} ln ∣ x ∣

\int \frac{1}{36 ( x - 6 )} d x = \frac{1}{36} ln ∣ x - 6 ∣

\int \frac{1}{6 ( x - 6 ) ^{2}} d x = - \frac{1}{6 ( x - 6 )}

= 6 (\frac{1}{36} ln ∣ x ∣ - \frac{1}{36} ln ∣ x - 6 ∣ - \frac{1}{6 ( x - 6 )})

Simplify

6 (\frac{1}{36} ln ∣ x ∣ - \frac{1}{36} ln ∣ x - 6 ∣ - \frac{1}{6 ( x - 6 )}) : \frac{1}{6} ln ∣ x ∣ - \frac{1}{6} ln ∣ x - 6 ∣ - \frac{1}{x - 6}

= \frac{1}{6} ln ∣ x ∣ - \frac{1}{6} ln ∣ x - 6 ∣ - \frac{1}{x - 6}

Add a constant to the solution

= \frac{1}{6} ln ∣ x ∣ - \frac{1}{6} ln ∣ x - 6 ∣ - \frac{1}{x - 6} + C

What is the integral of 6/(x(x-6)^2) ?
The integral of 6/(x(x-6)^2) is 1/6 ln|x|-1/6 ln|x-6|-1/(x-6)+C