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

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

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

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

- \frac{ln ( 2 x ^{2} + 1 )}{x} + 22 arctan (2 x) + C

Solution steps

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

Apply Integration By Parts

= - \frac{ln ( 2 x ^{2} + 1 )}{x} - \int - \frac{4}{2 x ^{2} + 1} d x

\int - \frac{4}{2 x ^{2} + 1} d x = - 22 arctan (2 x)

= - \frac{ln ( 2 x ^{2} + 1 )}{x} - (- 22 arctan (2 x))

Simplify

= - \frac{ln ( 2 x ^{2} + 1 )}{x} + 22 arctan (2 x)

Add a constant to the solution

= - \frac{ln ( 2 x ^{2} + 1 )}{x} + 22 arctan (2 x) + C

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