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How should I evaluate this definite integral? I am unable to figure out how to start.

$$\int \tan^{-1} \left(1 + x + x^{2}\right) dx $$

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Integration by parts... –  Artem Dec 17 '12 at 3:09
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Proceed by integration by parts, we get \begin{align*} \int \arctan \left( 1 + x + x^2 \right) dx & = x \arctan \left(1+x+x^2 \right) - \int \dfrac{x(2x+1)}{(x^2+x+1)^2+1} dx \end{align*} Now note that $$\left(x^2+x+1 \right)^2+1 = \left(x^2+1 \right) \left(x^2+2x+2 \right)$$ Hence, $$\dfrac{x(2x+1)}{(x^2+x+1)^2+1} = \dfrac{x}{x^2+1} - \dfrac{x}{x^2 + 2x+2}$$ Hence, $$\int \dfrac{x(2x+1)}{(x^2+x+1)^2+1} dx = \int \dfrac{x dx}{x^2+1} - \int \dfrac{x dx}{x^2 + 2x+2}$$ $$\int \dfrac{x dx}{x^2+1} = \dfrac12 \log \left(1+x^2 \right)$$ $$\int \dfrac{x dx}{x^2 + 2x+2} = \int \dfrac{\left(x + 1 \right) dx}{\left( x+1 \right)^2+1} - \int \dfrac{dx}{\left( x+1 \right)^2+1} = \dfrac12 \log \left((x+1)^2+1\right) - \arctan(x+1)$$ Putting all this together, we get that $$x \arctan \left( 1+x+x^2 \right) - \dfrac12 \log \left( 1+x^2\right) + \dfrac12 \log \left((x+1)^2+1\right) - \arctan(x+1)$$ Rearranging gives us $$x \arctan \left( 1+x+x^2 \right) - \arctan(x+1) + \dfrac12 \log \left(\dfrac{(x+1)^2+1}{x^2+1}\right) + \text{ constant}$$

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plus a constant! –  Artem Dec 17 '12 at 3:22
    
@Artem Yes. I have added. Thanks. –  user17762 Dec 17 '12 at 3:22
    
@Artem For some reason the link is not accessible. –  user17762 Dec 17 '12 at 3:24
    
I edited the link. And of course that was just a joke. –  Artem Dec 17 '12 at 3:24
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