What is the integral of log_{e}(x^2+1) ?

The integral of log_{e}(x^2+1) is xln(x^2+1)-2(-arctan(x)+x)+C

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integral of log_{e}(x^2+1)

\int lo g_{e} (x^{2} + 1) d x

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

Solution steps

\int lo g_{e} (x^{2} + 1) d x

Apply log rule:

lo g_{e} (x) = ln (x)

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

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

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What is the integral of log_{e}(x^2+1) ?
The integral of log_{e}(x^2+1) is xln(x^2+1)-2(-arctan(x)+x)+C