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

The integral of (3x-1)/(x^2+x) is 4ln|x+1|-ln|x|+C

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

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

4 ln ∣ x + 1 ∣ - ln ∣ x ∣ + C

Solution steps

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

Expand

\frac{3 x - 1}{x ^{2} + x} : \frac{3 x}{x ^{2} + x} - \frac{1}{x ^{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{3 x}{x ^{2} + x} d x - \int \frac{1}{x ^{2} + x} d x

\int \frac{3 x}{x ^{2} + x} d x = 3 ln ∣ x + 1 ∣

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

= 3 ln ∣ x + 1 ∣ - (ln ∣ x ∣ - ln ∣ x + 1 ∣)

Simplify

= 4 ln ∣ x + 1 ∣ - ln ∣ x ∣

Add a constant to the solution

= 4 ln ∣ x + 1 ∣ - ln ∣ x ∣ + C

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What is the integral of (3x-1)/(x^2+x) ?
The integral of (3x-1)/(x^2+x) is 4ln|x+1|-ln|x|+C