Question

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

• A. $\frac{\pi }{2}-\log 2$
• B. $\log 2$
• C. $\frac{\pi }{2}+\frac{1}{2}\log 2$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (A)

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

Consider, $I={\int }_0^1{cot}^{-1}(1-x+x^2)dx$

$\Rightarrow$ $I={\left[x{cot}^{-1}(1-x+x^2)\right]}_0^1+{\int }_0^1\frac{x(2x-1)}{{(1-x+x^2)}^2+1}dx$

$\Rightarrow$ $I=\frac{\pi }{4}+\frac{1}{2}{\int }_0^1\frac{2x-2}{x^2-2x+2}dx+{\int }_0^1\frac{1}{x^2-2x+2}dx-{\int }_0^1\frac{x}{x^2+1}dx$

$\Rightarrow$ $I=\frac{\pi }{4}+{\left[\frac{\log (x^2-2x+2)}{2}\right]}_0^1+{\left[{tan}^{-1}(x-1)\right]}_0^1-{\left[\frac{\log (x^2+1)}{2}\right]}_0^1$

$\Rightarrow$ $I=\frac{\pi }{4}-\frac{\log 2}{2}+0+\frac{\pi }{4}-\frac{\log 2}{2}=\frac{\pi }{2}-\log 2$