Question

Choose the correct answer.

$\text{The value of}{\int }_0^{1}ta{n}^{-1}\left(\frac{2x-1}{1+x-x^2}\right)dxis\left(a\right)1\left(b\right)zero\left(c\right)-1\left(d\right)\frac{\pi }{4}$

Answer 1

$LetI={\int }_0^{1}tan{t}^{-1}\left(\frac{2x-1}{1+x-x^2}\right)dx={\int }_0^{1}tan{t}^{-1}\left(\frac{x+(x-1)}{1-x(x-1)}\right)dx{\int }_0^1\{ta{n}^{-1}x+ta{n}^{-1}(x-1\left)\right\}dx[\because ta{n}^{-1}A+ta{n}^{-1}B=ta{n}^{-1}\left(\frac{A+B}{1-AB}\right)]\Rightarrow I={\int }_0^1\{ta{n}^{-1}x-ta{n}^{-1}(1-x\left)\right\}dx..\left(i\right)Also,I={\int }_0^1\left\{ta{n}^{-1}\right(1-x)-ta{n}^{-1}(1-(1-x)\left)\right\}dx[\because {\int }_0^{a}f(x)dx-{\int }_0^{a}f(a-x\left)dx\right]\Rightarrow I={\int }_0^1\left[ta{n}^{-1}\right(1-x)-ta{n}^{-1}(x\left)\right]dx...\left(ii\right)$

On adding Eqs. (i) and (ii), we get

2I = 0 $\Rightarrow$ I = 0. Hence, option (b) is correct.