The value of the integral _0^1x^-1(1-x^2+x^4)dx is:

Question

The value of the integral $\displaystyle\int_0^1x\cot^{-1}\left(1-x^2+x^4\right)dx$ is:

Accepted Answer

$$\int_0^1x\cot^{-1}(1-x^2+x^4)\,dx=\int_0^1x\tan^{-1}\left(\frac1{1+x^4-x^2}\right)dx=\int_0^1x\tan^{-1}\left(\frac{x^2-(x^2-1)}{1+x^2(x^2-1)}\right)dx$$

$$=\int_0^1x\left[\tan^{-1}x^2-\tan^{-1}(x^2-1)\right]dx$$

Substituting $t=x^2$ in the first part and $k=x^2-1$ in the second (both with $x\,dx=\tfrac12dt$):
$$=\frac12\int_0^1\tan^{-1}t\,dt-\frac12\int_{-1}^0\tan^{-1}k\,dk$$

Using $\displaystyle\int\tan^{-1}u\,du=u\tan^{-1}u-\tfrac12\ln(1+u^2)$:
$$=\frac12\left[\frac\pi4-\frac12\ln2\right]-\frac12\left[-\frac\pi4-\frac12\ln2\right]=\frac\pi4-\frac12\ln2$$

Solutions · 1