Solve Indefinite Integral: $\int\frac{\arctan x}{1+x^2}dx$

[paulmdrdo1]

how would i go about solving this

$\displaystyle\int\frac{\arctan x}{1+x^2}dx$?

i tried substitution but i didn't work.

 

[HallsofIvy]

That's very strange. Did you try the obvious substitution $$u= tan^{-1}(x)$$. What is the derivative of $$tan^{1}(x)$$? (And do you understand that $$tan^{-1}(x)$$ here is the arctangent, NOT the $$\frac{1}{tan(x)}$$?!)

 

[paulmdrdo1]

yes, i know that. but how do i use substitution here?

 

[Ackbach]

Following HoI's suggestion, we get that $u=\tan^{-1}(x)$, and
$$du= \frac{1}{1+x^{2}} \, dx,$$
and the integral becomes
$$\int u \, du.$$
Can you continue?

 

[paulmdrdo1]

this is my answer,

$\displaystyle\frac{1}{2}(\tan^{-1}x)^2+C$

 

[Plato2]

this is my answer,

$\displaystyle\frac{1}{2}(Tan^{-1})^2+C$

That is the correct idea, but awful notation: $\displaystyle\frac{1}{2}(\arctan^{2}(x))+C$

 

[topsquark]

That is the correct idea, but awful notation: $\displaystyle\frac{1}{2}(\arctan^{2}(x))+C$

I find that "atn(x)" works quite well also.

-Dan