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

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SUMMARY

The integral $\int\frac{\arctan x}{1+x^2}dx$ can be solved using the substitution $u=\tan^{-1}(x)$, leading to $du= \frac{1}{1+x^{2}} \, dx$. This transforms the integral into $\int u \, du$, which evaluates to $\frac{1}{2}(\arctan^{2}(x)) + C$. The notation for the final answer should be clearly presented as $\frac{1}{2}(\arctan^{2}(x)) + C$ for clarity and correctness.

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paulmdrdo1
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how would i go about solving this

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

i tried substitution but i didn't work.
 
Last edited:
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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)}?!)
 
yes, i know that. but how do i use substitution here?
 
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?
 
this is my answer,

$\displaystyle\frac{1}{2}(\tan^{-1}x)^2+C$
 
Last edited:
paulmdrdo said:
this is my answer,

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

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

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