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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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