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I do not know how to integrate

$$\int{\arctan{x} \over 1+x^2}dx$$

The answer is given to be: $\arctan{x} + 1/2 \ln{|1+x^2|}+C$

I'd appreciate your help very much. Thanks!

Edit 1: it must be done with substitution only.

Edit 2: Yes. It's not the answer. Thank you all. I'll delete this in a minute. Thanks!!

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    $\begingroup$ If I am reading the question right, I have trouble with the given answer. Let $u=\arctan x$. $\endgroup$
    – André Nicolas
    Sep 17, 2015 at 5:27
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    $\begingroup$ If the integrand is correct, the given answer is wrong ! $\endgroup$
    – Claude Leibovici
    Sep 17, 2015 at 5:31
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    $\begingroup$ Hi and welcome to Math.SE. Please share your thoughts on the problem. The answer given is clearly wrong (just differentiate it, and you'll see). Is the problem written correctly as it is now, when edited? Also, in the future, you could write your questions using mathjax. $\endgroup$
    – mickep
    Sep 17, 2015 at 5:33
  • $\begingroup$ Could it be? My teacher gave me that answer. Maybe you're right, he's wrong. $\endgroup$
    – Mylane
    Sep 17, 2015 at 5:33
  • $\begingroup$ This is a related question, but about a definite integral: What is the value of $\int_0^1 \frac{\arctan x}{1+x^{2}} dx$? $\endgroup$
    – Martin Sleziak
    Sep 17, 2015 at 16:11

2 Answers 2

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If you set $u=\arctan{x}$ you get $du={dx\over 1+x^2}$ and the integral rewrites as

$$\int udu={u^2\over 2}+C$$

And I share the trouble of Andre Nicolas and Claude Leibovici and it has nothing to do with the fact that the three of us are French

If you derive the "answer" that's given you get

$${1+x\over 1+x^2}\neq {\arctan{x}\over 1+x^2}$$

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  • $\begingroup$ Perhaps French is forever, but was French is more accurate. $\endgroup$
    – André Nicolas
    Sep 17, 2015 at 5:54
  • $\begingroup$ Could anyone suspect a French coalition on this site ? Cheers :-) $\endgroup$
    – Claude Leibovici
    Sep 17, 2015 at 5:54
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Notice that $ff'$ is the derivative of $\frac12 f^2+C$ and in your example $f(x)=\arctan x$.

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