# Solving of an integral

$$\int \left(\frac{x^2 + \arctan(x)}{1 + x^2}\right) dx$$

Could anyone help me calculate this integral? Thanks in advance.

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Break the integral in two :

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

To solve the second one : Notice $d( \arctan x ) = \frac{ dx}{1 + x^2}$ So

$$\int \frac{ \arctan x }{1 + x^2} dx = \int \arctan x\, d(\arctan x ) = (\arctan x)^2/2 + K$$

As for the first integral, notice

$$\int \frac{ x^2}{1 + x^2} = \int \frac{ x^2 + 1 - 1 }{1+x^2} dx = \int dx - \int \frac{ 1}{1 + x^2 } dx = x - \arctan x$$

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nice, :) +1, good answer – Rustyn Apr 8 '14 at 3:30
For $\arctan x$ you can use \arctan x in $\LaTeX$ – Ross Millikan Apr 8 '14 at 3:31
Thanks for the answer.. I have one question, I don't understand how you solved the second one? Can you explain in depth how it's done? – user133022 Apr 8 '14 at 3:32
the trick is adding one and substracting one. IT is a trick. – ProbabilityGuy Apr 8 '14 at 3:33
I meant the second part, not the first one. – user133022 Apr 8 '14 at 3:34

Here's another way: $$\int \left(\frac{x^2 + \arctan(x)}{1 + x^2}\right) dx=\int \left(x^2 + \arctan(x)\right) d(\arctan x)=\\ \left(x^2 + \arctan(x)\right)\arctan x-\int \arctan xd(x^2 + \arctan(x))=\\ \left(x^2 + \arctan(x)\right)\arctan x-\int \arctan xd(x^2)-\int \arctan xd(\arctan(x))=\\ \left(x^2 + \arctan(x)\right)\arctan x-\dfrac{(\arctan x)^2}{2}-\int \arctan xd(x^2)$$

$$\text{Now, }\int \arctan xd(x^2)=x^2\arctan x-\int \dfrac{x^2}{x^2+1}dx$$

$$\text{This gives }\dfrac{(\arctan x)^2}{2}+\int \dfrac{x^2+1-1}{x^2+1}dx=\dfrac{(\arctan x)^2}{2}+x-\arctan x+C\\ \implies \int \left(\frac{x^2 + \arctan(x)}{1 + x^2}\right) dx=\dfrac{(\arctan x)^2}{2}+x-\arctan x+C$$

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