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

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

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

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

- ln ∣ x ∣ + 2 ln ∣ x - 1 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{2}{x - 1} d x = 2 ln ∣ x - 1 ∣

= - ln ∣ x ∣ + 2 ln ∣ x - 1 ∣

Add a constant to the solution

= - ln ∣ x ∣ + 2 ln ∣ x - 1 ∣ + C

\int 6 x sin (x) cos (x) d x

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

y = cos^{2} (e^{c o s^{2} (t)})

\int \frac{( 1 + x )}{1 - x} d x

49 y^{^{''}} + 126 y^{^{'}} + 97 y = 0

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