Question

${\int }_0^1{\cot }^{-1}(1-x+x^2)dx$ equals

• A. $\frac{\pi }{2}+\log 2$
• B. $\frac{\pi }{2}-\log 2$
• C. $\pi -\log 2$
• D. None of these

Answer 1

Answer: (B)

$\frac{\pi }{2}-\log 2$

$I{=}_0^1\int co{t}^{-1}(1-x+x^2)dx$

${=}_0^1\int ta{n}^{-1}\left(\frac{1}{1-x+x^2}\right)dx$

${=}_0^1\int ta{n}^{-1}\left(\frac{1}{1-x(1-x)}\right)dx$

${=}_0^1\int [ta{n}^{-1}+ta{n}^{-1}(1-x\left)\right]dx$

${=}_0^1\int ta{n}^{-1}x{+}_0^1\int ta{n}^{-1}(1-x)dx$

${=}_0^1\int ta{n}^{-1}x{+}_0^1\int ta{n}^{-1}(1-x)dx$

$=2_0^1\int ta{n}^{-1}xdx$

$=2[x+ta{n}^{-1}x{]}_0^1{-}_0^1\int (\frac{2x}{1+x^2})dx$

$=2.\frac{\pi }{4}-(log1+x^2{)}_0^1$

$=\frac{\pi }{2}-(log2-log1)$

$=\frac{\pi }{2}-log2.$