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I need to evaluate
$$\int_0^1\arctan(1-x+x^2)dx$$ What I did: First I assume $$I=\int_0^1\arctan(1-x+x^2)dx=\int_0^1\arctan((x-\frac{1}{2})^2+\frac{3}{4})dx$$ Since the function is symmetric about $\frac{1}{2}$, as $f(\frac{1}{2}+t)=f(\frac{1}{2}-t)$, $$I=2\int_0^{\frac{1}{2}}\arctan((x-\frac{1}{2})^2+\frac{3}{4})dx$$ Since $\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$, I get $I$ as $$I=2\int_0^{\frac{1}{2}}\arctan(x^2+\frac{3}{4})dx$$ The integration by parts method, here seems hectic. Does anyone know how to evaluate this integral, in a different way, or ahead of whatever point I've reached?

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2 Answers

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$$\begin{align} \int_0^1\tan^{-1}(1-x+x^2)dx &= \frac{\pi}{2}-\int_{0}^{1}\tan^{-1}\left(\frac1{1-x+x^2}\right)dx \\ &=\frac{\pi}{2}-\int_0^1 \left(\tan^{-1}(x)-\tan^{-1}(x-1)\right)dx \end{align}$$That should simplify everything.

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You can use this technique, using trigonometric formulae to evaluate this integral easily.

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