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

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

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

\int \frac{ln ( 1 + 3 x )}{x ^{2}} d x

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

Solution steps

\int \frac{ln ( 1 + 3 x )}{x ^{2}} d x

Apply Integration By Parts

: - \frac{ln ( 1 + 3 x )}{x} - \int - \frac{3}{x ( 1 + 3 x )} d x

= - \frac{ln ( 1 + 3 x )}{x} - \int - \frac{3}{x ( 1 + 3 x )} d x

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

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

Simplify

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

Add a constant to the solution

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

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