Question

\[\huge \int \frac{arctan(x)}{1+x^2} dx \]

Answer 1

put \(x= tan \theta\)
\[dx= \sec^2 \theta\]
we get
\[\int \frac { \theta \sec^2 \theta}{sec^2 \theta} d\theta\]
we have now
\[\int \theta d\theta= \frac{{\theta}^2}{2} +c\]
substitute \(\theta = \tan ^{-1} x\)

Answer 2

let u = arctan(x) then du/dx = 1/(1+x^2) or du = 1/(1+x^2) dx

so the problem is now

\[\int\limits u du\]

\[= 1/2 u^2 + c\]

= 1/2 (arctan(x))^2 + c

Answer 3

campbell is right

Answer 4

got the same

Answer 5

oops it should read \[u^2/2 + c\]