Question

${\int }_0^1{\tan }^{-1}\left(\frac{2x-1}{1+x-x^2}\right)dx$ is equal to

• A. $0$
• B. $1$
• C. $-1$
• D. None of these

Answer 1

The correct option is A $0$

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

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

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

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

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

$[xta{n}^{-1}x{]}_0^1-{\int }_0^1\frac{x}{1+x^2}dx$

$1+x^2=t$

$2xdx=dt$

$\left[1ta{n}^{-1}\right]-\frac{1}{2}{\int }_1^2\frac{dt}{t}$

$\frac{\pi }{4}-\frac{1}{2}ln2$

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

$1-x=t$

$-dx=dt$

${\int }_0^{1}ta{n}^{-1}\left(t\right)dt$

$[tta{n}^{-1}t{]}_0^1-{\int }_0^1\frac{t}{1+t^2}dt$

$=\frac{\pi }{4}-\frac{1}{2}ln2$

$e{q}^n-\left(1\right)\Rightarrow$

$\frac{\pi }{4}-\frac{1}{2}ln-\frac{\pi }{4}+\frac{1}{2}ln2$

$0$