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

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

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

integral

\frac{1}{x ^{2} ( x - 1 )}

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

Simplify

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

Add a constant to the solution

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

\frac{x ^{2}}{( x + 2 )}

cos (2 x + y)

\frac{1}{n ^{2}}

3^{2} x

\frac{( x ^{3} - 3 x )}{( 1 - x ^{4} )}

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