Finding $k$ such that $\int_{0}^{1} \text{arccot}\bigl(1-x+x^{2}\bigr) \ dx = k \cdot\int_{0}^{1} \arctan(x) \ dx$ Suppose $$\int_{0}^{1} \text{arccot}\bigl(1-x+x^{2}\bigr) \ dx = k \cdot \int_{0}^{1} \arctan(x) \ dx$$ then find the value of $k$.
I know $\text{arccot}(x) = \arctan \left(\frac{1}{x} \right)$. But after that what to do ?:(

I guess you can proceed as follows:
\begin{align*}
\int_{0}^{1} \cot^{-1}(1-x+x^{2}) \ dx &= \int_{0}^{1} \tan^{-1}\biggl(\frac{1}{1-x+x^{2}}\biggr) \ dx \\ &= \int_{0}^{1} \tan^{-1}\biggl(\frac{(1-x)+x}{1-x\cdot(1-x)}\biggr) \ dx \\&=\int_{0}^{1} \tan^{-1}{(1-x)} \ dx + \int_{0}^{1} \tan^{-1}(x) \ dx \\\ &= 2 \cdot \int_{0}^{1} \tan^{-1}(x) \ dx \qquad \Bigl[ \because \small \int_{0}^{a} f(x) \ dx = \int_{0}^{a} f(a-x) \ dx \Bigr]
\end{align*}
