int0^1 tan^(-1) ((2x-1)/(1+x-x^2)) dx is...

`int_0^1 tan^(-1) ((2x-1)/(1+x-x^2)) dx` is equal to

A

`1//4`

B

`1//2`

C

`1`

D

`2`

Text Solution

Verified by Experts

The correct Answer is:

B

`I=int_(0)^(1)("tan"^(-1)(x/(x+1)))/("tan"^(-1)((1+2x-2x^(2))/2))dx`…………….1
`=int_(0)^(1)("tan"^(-1)((1-x)/((1-x)+1)))/("tan"^(-1)((1+2(1-x)-2(1-x)^(2))2))dx`
`=int_(0)^(1)("tan"^(-1)((1-x)/(2-x)))/("tan"^(-1)((1+2x-2x^(2))/2))dx`……….2
ADding 1 and 2 we get
`2I=int_(0)^(1)("tan"^(-1)(x/(x+1))+"tan"^(-1)((1-x)/(2-x)))/("tan"^(-1)((1+2x-2x^(2))/2))dx`
`=int_(0)^(1)("tan"^(-1)((x/(x+1)+(1-x)/(2-x))/(1-x/(x+1) . (1-x)/(2-x))))/("tan"^(-1)((1+2x-2x^(2))/2))dx`
`=int_(0)^(1)((1+2x-2x^(2))/(2))/("tan"^(-1)((1+2x-2x^(2))/2))dx=int_(0)^(1) 1dx=1`
`implies I=1//2`

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