What is the integral of 3/(x^2(x-1)) ?

The integral of 3/(x^2(x-1)) is 3(-ln|x|+1/x+ln|x-1|)+C

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integral of 3/(x^2(x-1))

\int \frac{3}{x ^{2} ( x - 1 )} d x

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

Solution steps

\int \frac{3}{x ^{2} ( x - 1 )} d x

Take the constant out:

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

= 3 \cdot \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}

= 3 \cdot \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

= 3 (- \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 ∣

= 3 (- ln ∣ x ∣ - (- \frac{1}{x}) + ln ∣ x - 1 ∣)

Simplify

= 3 (- ln ∣ x ∣ + \frac{1}{x} + ln ∣ x - 1 ∣)

Add a constant to the solution

= 3 (- ln ∣ x ∣ + \frac{1}{x} + ln ∣ x - 1 ∣) + C

\frac{\partial}{\partial x} (y^{2} ln (x))

\int \frac{x ^{6} - 5 x}{x ^{4}} d x

\frac{d}{d x} (e^{2} (sin (\frac{π}{4}) - 1))

\frac{s - 1}{2 s ^{2} + 8 s + 11}

1 + 5 x

What is the integral of 3/(x^2(x-1)) ?
The integral of 3/(x^2(x-1)) is 3(-ln|x|+1/x+ln|x-1|)+C