Question

The value of `int_0^1 tan^-1 ((2x - 1)/(1 + x - x^2))  dx` is

विकल्प

• 1

• 0

• – 1

• `pi/4`

Answer 1

0

Explanation:

`tan^-1 ((2x - 1)/(1 + x - x^2)) = tan^-1 [(x + (x - 1))/(1 - x(x - 1))]`

= `tan^-1x + tan^-1 (x - 1)`  ......(1)

`[because tan^-1x + tan^-1y = tan^-1  ((x + y)/(1 - xy))]`

∴ Let, I = `int_0^1 tan^-1 ((2x - 1)/(1 + x - x^2))  dx`

= `[int_0^1  tan^-1x + tan^-1 (x - 1)]`

= `[because int_0^a f(x)  dx = int_0^a  f(a - x)]  dx`

= `int_0^1 [- tan^-1 (x - 1) - tan^-1 x]  dx`

∴ I = `- int_0^1 [tan^-1 x + tan^-1 (x - 1)]  dx`  ......(2)

Adding (1) and (2)

2I = 0 or I = 0