What is the integral of 5/(x(x-5)^2) ?

The integral of 5/(x(x-5)^2) is 1/5 ln|x|-1/5 ln|x-5|-1/(x-5)+C

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integral of 5/(x(x-5)^2)

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

\frac{1}{5} ln ∣ x ∣ - \frac{1}{5} ln ∣ x - 5 ∣ - \frac{1}{x - 5} + C

Solution steps

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

Take the constant out:

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

= 5 \cdot \int \frac{1}{x ( x - 5 ) ^{2}} d x

Take the partial fraction of

\frac{1}{x ( x - 5 ) ^{2}} : \frac{1}{25 x} - \frac{1}{25 ( x - 5 )} + \frac{1}{5 ( x - 5 ) ^{2}}

= 5 \cdot \int \frac{1}{25 x} - \frac{1}{25 ( x - 5 )} + \frac{1}{5 ( x - 5 ) ^{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

= 5 (\int \frac{1}{25 x} d x - \int \frac{1}{25 ( x - 5 )} d x + \int \frac{1}{5 ( x - 5 ) ^{2}} d x)

\int \frac{1}{25 x} d x = \frac{1}{25} ln ∣ x ∣

\int \frac{1}{25 ( x - 5 )} d x = \frac{1}{25} ln ∣ x - 5 ∣

\int \frac{1}{5 ( x - 5 ) ^{2}} d x = - \frac{1}{5 ( x - 5 )}

= 5 (\frac{1}{25} ln ∣ x ∣ - \frac{1}{25} ln ∣ x - 5 ∣ - \frac{1}{5 ( x - 5 )})

Simplify

5 (\frac{1}{25} ln ∣ x ∣ - \frac{1}{25} ln ∣ x - 5 ∣ - \frac{1}{5 ( x - 5 )}) : \frac{1}{5} ln ∣ x ∣ - \frac{1}{5} ln ∣ x - 5 ∣ - \frac{1}{x - 5}

= \frac{1}{5} ln ∣ x ∣ - \frac{1}{5} ln ∣ x - 5 ∣ - \frac{1}{x - 5}

Add a constant to the solution

= \frac{1}{5} ln ∣ x ∣ - \frac{1}{5} ln ∣ x - 5 ∣ - \frac{1}{x - 5} + C

What is the integral of 5/(x(x-5)^2) ?
The integral of 5/(x(x-5)^2) is 1/5 ln|x|-1/5 ln|x-5|-1/(x-5)+C