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

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

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

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

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

Solution steps

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

Apply log rule:

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

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

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

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