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Find integral of $$\int\limits^{\frac{{\pi}}{2}}_{{0}} \operatorname{arccot}\left(1-x+x^2\right)\,dx $$

More specifically,

Show that $$\displaystyle\int\limits^{\frac{{\pi}}{2}}_{{0}} \operatorname{arccot}\left(1-x+x^2\right)\,dx = \frac{\pi}{2}- \log(2)$$

I started out with integration by parts since I could not make a efficient substitution. I now have $-x\operatorname{arccot}\left(x^2+x-1\right)-{\displaystyle\int}\dfrac{\left(2x+1\right)x}{\left(x^2+x-1\right)^2+1}\,\mathrm{d}x$, which just seems more complex. Any hints or ideas would be greatly appreciated.

Working:

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  • $\begingroup$ for the second component of your last step, you can expand the integral by partial fraction decomposition. You will now find $\int\bigg(-\dfrac{2x^2}{(x^2+x-1)^2+1} - \dfrac{x}{(x^2+x-1)+1} \bigg) \textrm{d}x$. However, I'm not sure how to proceed, since factoring the denominator looks difficult. $\endgroup$ – Rice4000 May 31 at 13:13
  • $\begingroup$ The solution looks also difficult. $\endgroup$ – Dr. Sonnhard Graubner May 31 at 13:14
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    $\begingroup$ Are you sure you typed it right? Why would anyone ever combine $\arctan $ with $\frac{\pi}{2}$ I bet the upper bound is $1$. $\endgroup$ – カカロット May 31 at 13:18
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    $\begingroup$ Note that $\displaystyle\int\limits^{\frac{{\pi}}{2}}_{{0}} \operatorname{arccot}\left(1-x-x^2\right)\,dx $$ \ne \frac{\pi}{2}- \log(2)$. This can be verified by wolfram alpha wolframalpha.com/input/… $\endgroup$ – Mathphile May 31 at 13:20
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    $\begingroup$ If the upper limit is 1 rather than $\pi/2$ then the result is correct. $\endgroup$ – user10354138 May 31 at 13:30
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$$\arctan\left(\frac{1}{1-x+x^2}\right)=\arctan(1-x)+\arctan x$$ But we don't need to evaluate two integrals since using the substitution $1-x=x$ there is $$\int_0^1 \arctan(1-x)dx=\int_0^1\arctan xdx$$ $$\Rightarrow \int_0^1 \text{arccot}(1-x+x^2)dx=2\int_0^1 \arctan xdx$$ $$=2x\arctan x|_0^1 -\int_0^1 \frac{2x}{1+x^2}dx=\frac{\pi}{2}-\ln 2$$

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Hint:$$\begin{align}\operatorname{arccot}\left(1-x+x^2\right)&=\arctan\left(\frac{1}{1-x(1-x)}\right)\\&=\arctan\left(\frac{1-x+x}{1-x(1+x)}\right)\\&=\arctan(1-x)+\arctan(x)\end{align}$$

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