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

The integral of ln(1+t^2) is tln(1+t^2)-2(-arctan(t)+t)+C

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integral of ln(1+t^2)

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

\int \frac{tan ^{3} ( ln ( x ) )}{3 x} d x

2 w^{3 π}, at 7

\frac{k}{s}

6 sec (x) tan (x)

n = 3 \sum \infty \frac{6}{n ( n - 2 )}

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