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Evaluate: int(0)^1tan^(-1)x/(1+x^2) dx...

Evaluate: `int_(0)^1tan^(-1)x/(1+x^2) dx`

Text Solution

Verified by Experts

The correct Answer is:
`2pi`
`int_(1)^(3)("tan"^(-1)x/(x^(2)+1)+"tan"^(-1)(x^(2)+1)/x)dx`
`=int_(-1)^(0)("tan"^(-1)x/(x^(2)+1)+"tan"^(-1)(x^(2)+1)/x)dx`
`+int_(0)^(3)("tan"^(-1)x/(x^(2)+1)+"tan"^(-1)(x^(2)+1)/x)dx`
`int_(-1)^(0)-(pi)/2 dx+int_(0)^(3)(pi)/2 dx`
`=[-(pi)/2 x]_(-1)^(0)+[(pi)/2x]_(0)^(3)`
`=2pi`
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