Question

`int_0^1 tan^-1 ((2x - 1)/(1 + x - x^2))` dx = ?

Options

• 2

• 1

• 0

• 4

Answer 1

0

Explanation:

We have,

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

I = `int_0^1 tan^-1 ((2 (1 - x) - 1)/(1 + (1 - x) - (1 - x)^2))`dx   ...`[because int_0^"a" "f"(x) "dx" = int_0^"a" "f"("a - x") "dx"]`

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

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

2I = 0

I = 0