What is the integral of (x+1)/(x^2-4x) ?

The integral of (x+1)/(x^2-4x) is -1/4 ln|x|+5/4 ln|x-4|+C

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integral of (x+1)/(x^2-4x)

\int \frac{x + 1}{x ^{2} - 4 x} d x

- \frac{1}{4} ln ∣ x ∣ + \frac{5}{4} ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{x + 1}{x ^{2} - 4 x} d x

Expand

\frac{x + 1}{x ^{2} - 4 x} : \frac{x}{x ^{2} - 4 x} + \frac{1}{x ^{2} - 4 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} - 4 x} d x + \int \frac{1}{x ^{2} - 4 x} d x

\int \frac{x}{x ^{2} - 4 x} d x = ln ∣ x - 4 ∣

\int \frac{1}{x ^{2} - 4 x} d x = - \frac{1}{4} ln ∣ x ∣ + \frac{1}{4} ln ∣ x - 4 ∣

= ln ∣ x - 4 ∣ - \frac{1}{4} ln ∣ x ∣ + \frac{1}{4} ln ∣ x - 4 ∣

Simplify

ln ∣ x - 4 ∣ - \frac{1}{4} ln ∣ x ∣ + \frac{1}{4} ln ∣ x - 4 ∣ : - \frac{1}{4} ln ∣ x ∣ + \frac{5}{4} ln ∣ x - 4 ∣

= - \frac{1}{4} ln ∣ x ∣ + \frac{5}{4} ln ∣ x - 4 ∣

Add a constant to the solution

= - \frac{1}{4} ln ∣ x ∣ + \frac{5}{4} ln ∣ x - 4 ∣ + C

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What is the integral of (x+1)/(x^2-4x) ?
The integral of (x+1)/(x^2-4x) is -1/4 ln|x|+5/4 ln|x-4|+C