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

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

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integral of ln(1+x^2)

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

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

Solution steps

\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

\frac{\partial}{\partial x} ((2 x + 8) ln (x y))

\int sin (20 x) cos (11 x) d x

f^{^{'}} (x) = 3 x + 2

x \to 3 lim (\frac{1}{( x - 3 ) ^{3}})

\frac{d}{d t} (arctan (4 t))

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