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

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

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

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

- \frac{1}{2} ln ∣ x ∣ + \frac{3}{2} ln ∣ x - 2 ∣ + C

Solution steps

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

Expand

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

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

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

= ln ∣ x - 2 ∣ - \frac{1}{2} ln ∣ x ∣ + \frac{1}{2} ln ∣ x - 2 ∣

Simplify

ln ∣ x - 2 ∣ - \frac{1}{2} ln ∣ x ∣ + \frac{1}{2} ln ∣ x - 2 ∣ : - \frac{1}{2} ln ∣ x ∣ + \frac{3}{2} ln ∣ x - 2 ∣

= - \frac{1}{2} ln ∣ x ∣ + \frac{3}{2} ln ∣ x - 2 ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ x ∣ + \frac{3}{2} ln ∣ x - 2 ∣ + C

\frac{\partial}{\partial s} (s \cdot cos (t))

f (x) = 6 cos (2 x)

f (x) = \frac{x ^{3} - 8}{x ^{2} + 3}

f (x) = 3 x^{2} + 2 x - 10, at x = 1

f (t) = e^{4 t s i n (2 t)}

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