If 2_0^1 ^-1x,dx = _0^1 ^-1(1-x+x^2),dx, then _0^1 ^-1(1-x+x^2),dx is equal to:

Question

If $2\displaystyle\int_0^1 	an^{-1}x\,dx = \int_0^1 \cot^{-1}(1-x+x^2)\,dx$, then $\displaystyle\int_0^1 	an^{-1}(1-x+x^2)\,dx$ is equal to:

Accepted Answer

Since $\cot^{-1}	heta = \dfrac{\pi}{2}-	an^{-1}	heta$:

$2\int_0^1 	an^{-1}x\,dx = \int_0^1 \dfrac{\pi}{2}\,dx - \int_0^1 	an^{-1}(1-x+x^2)\,dx$

$\int_0^1 	an^{-1}(1-x+x^2)\,dx = \dfrac{\pi}{2} - 2\int_0^1	an^{-1}x\,dx$ (a)

Let $I_1 = \int_0^1 	an^{-1}x\,dx = [(	an^{-1}x)x]_0^1 - \int_0^1 \dfrac{x}{1+x^2}\,dx = \dfrac{\pi}{4} - \dfrac12\log 2$

By (a): $\dfrac{\pi}{2} - 2\left(\dfrac{\pi}{4}-\dfrac12\log2ight) = \log 2$

Solutions · 1