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

  • A
    $\frac{\pi}{2} + \log 2$
  • B
    $\log 2$
  • C
    $\frac{\pi}{2} - \log 4$
  • D
    $\log 4$

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Similar Questions

Let $g_i: \left[\frac{\pi}{8}, \frac{3\pi}{8}\right] \rightarrow \mathbb{R}, i=1, 2$,and $f: \left[\frac{\pi}{8}, \frac{3\pi}{8}\right] \rightarrow \mathbb{R}$ be functions such that $g_1(x)=1, g_2(x)=|4x-\pi|$ and $f(x)=\sin^2 x$,for all $x \in \left[\frac{\pi}{8}, \frac{3\pi}{8}\right]$.
Define $S_i = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} f(x) \cdot g_i(x) dx, i=1, 2$.
$(1)$ The value of $\frac{16S_1}{\pi}$ is.
$(2)$ The value of $\frac{48S_2}{\pi^2}$ is.

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