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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Take the partial fraction of

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

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

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

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

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

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

= 2 (- \frac{1}{25} ln ∣ x ∣ - (- \frac{1}{5 x}) + \frac{1}{25} ln ∣ x - 5 ∣)

Simplify

= 2 (- \frac{1}{25} ln ∣ x ∣ + \frac{1}{5 x} + \frac{1}{25} ln ∣ x - 5 ∣)

Add a constant to the solution

= 2 (- \frac{1}{25} ln ∣ x ∣ + \frac{1}{5 x} + \frac{1}{25} ln ∣ x - 5 ∣) + C

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