# 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 ?:(

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Wolfram Alpha says $k=2$ ; don't ask why to me, I don't know. I'm just saying perhaps it could help to know the answer. –  Patrick Da Silva Jun 2 '12 at 18:43

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*}

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Neat. Thanks for the help. –  shukla Jun 2 '12 at 18:49
All you need is $$\text{arccot}(1-x+x^2) = \arctan\left( \dfrac{x-(x-1)}{1+x(x-1)}\right) = \arctan(x) - \arctan(x-1)$$ Note that $$\displaystyle \int_0^1 \arctan(x-1) dx = - \int_0^1 \arctan(x) dx$$ with an appropriate change of variable i.e. $y=x-1$.
Hence, $$\int_0^1 \text{arccot}(1-x+x^2) dx = \int_0^1 \arctan(x) dx - \int_0^1 \arctan(x-1) dx = 2 \int_0^1 \arctan(x) dx$$ Hence, $k=2$.