# Find $\int^{\pi/2}_{{0}} \operatorname{arccot}(1-x+x^2)\,dx$

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:

• 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. – Rice4000 May 31 at 13:13
• The solution looks also difficult. – Dr. Sonnhard Graubner May 31 at 13:14
• Are you sure you typed it right? Why would anyone ever combine $\arctan$ with $\frac{\pi}{2}$ I bet the upper bound is $1$. – Zacky May 31 at 13:18
• 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/… – Mathphile May 31 at 13:20
• If the upper limit is 1 rather than $\pi/2$ then the result is correct. – user10354138 May 31 at 13:30

$$\arctan\left(\frac{1}{1-x+x^2}\right)=\arctan(1-x)+\arctan x$$ But we don't need to evaluate two integrals since using the substitution $$1-x=x$$ there is $$\int_0^1 \arctan(1-x)dx=\int_0^1\arctan xdx$$ $$\Rightarrow \int_0^1 \text{arccot}(1-x+x^2)dx=2\int_0^1 \arctan xdx$$ $$=2x\arctan x|_0^1 -\int_0^1 \frac{2x}{1+x^2}dx=\frac{\pi}{2}-\ln 2$$
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}