What is the integral of (7x+1)/(3x^2+6x) ?

The integral of (7x+1)/(3x^2+6x) is 13/6 ln|x+2|+1/6 ln|x|+C

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integral of (7x+1)/(3x^2+6x)

\int \frac{7 x + 1}{3 x ^{2} + 6 x} d x

\frac{13}{6} ln ∣ x + 2 ∣ + \frac{1}{6} ln ∣ x ∣ + C

Solution steps

\int \frac{7 x + 1}{3 x ^{2} + 6 x} d x

Expand

\frac{7 x + 1}{3 x ^{2} + 6 x} : \frac{7 x}{3 x ^{2} + 6 x} + \frac{1}{3 x ^{2} + 6 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{7 x}{3 x ^{2} + 6 x} d x + \int \frac{1}{3 x ^{2} + 6 x} d x

\int \frac{7 x}{3 x ^{2} + 6 x} d x = \frac{7}{3} ln ∣ x + 2 ∣

\int \frac{1}{3 x ^{2} + 6 x} d x = - \frac{1}{6} (ln ∣ x + 2 ∣ - ln ∣ x ∣)

= \frac{7}{3} ln ∣ x + 2 ∣ - \frac{1}{6} (ln ∣ x + 2 ∣ - ln ∣ x ∣)

Simplify

\frac{7}{3} ln ∣ x + 2 ∣ - \frac{1}{6} (ln ∣ x + 2 ∣ - ln ∣ x ∣) : \frac{13}{6} ln ∣ x + 2 ∣ + \frac{1}{6} ln ∣ x ∣

= \frac{13}{6} ln ∣ x + 2 ∣ + \frac{1}{6} ln ∣ x ∣

Add a constant to the solution

= \frac{13}{6} ln ∣ x + 2 ∣ + \frac{1}{6} ln ∣ x ∣ + C

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What is the integral of (7x+1)/(3x^2+6x) ?
The integral of (7x+1)/(3x^2+6x) is 13/6 ln|x+2|+1/6 ln|x|+C