What is the integral of 1/(x(x-5)^2) ?

The integral of 1/(x(x-5)^2) is 1/25 ln|x|-1/25 ln|x-5|-1/(5(x-5))+C

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integral of 1/(x(x-5)^2)

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

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

Solution steps

\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}}

= \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

= \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 )}

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

Add a constant to the solution

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

\frac{d}{d x} (ln (\frac{1}{x} + \frac{1}{x ^{2}}))

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

\int (2 x - 1) (3 x + 4) d x

\int 0.002 d x

x^{2} - 6 x + 5

What is the integral of 1/(x(x-5)^2) ?
The integral of 1/(x(x-5)^2) is 1/25 ln|x|-1/25 ln|x-5|-1/(5(x-5))+C