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

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

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integral of ln(3x^2+1)

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

x ln (3 x^{2} + 1) + \frac{2}{3} arctan (3 x) - 2 x + C

Solution steps

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

Apply Integration By Parts

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

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

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

= x ln (3 x^{2} + 1) - (- \frac{2}{3} arctan (3 x) + 2 x)

Simplify

= x ln (3 x^{2} + 1) + \frac{2}{3} arctan (3 x) - 2 x

Add a constant to the solution

= x ln (3 x^{2} + 1) + \frac{2}{3} arctan (3 x) - 2 x + C

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