1.

`Ifint_0^1cot^(-1)(1-x+x^2)dx=lambdaint_0^1tan^(-1)x dx ,t h e nlambdai se q u a lto`1 (b)2 (c) 3(d) 4A. `pi- log2`B. `pi+log2`C. `pi/2 +log2`D. `pi/2 - log2`

Answer» Correct Answer - D
Let `l=int_(0)^(1)cot^(-1)(1+x+x^(2))dx`
`=int_(0)^(1)tan^(-1)((1)/(x^(2)-x+1))dx`
`=int_(0)^(1)tan^(-1)((x-(x-1))/(1+x(x-1)))dx`
`=int_(0)^(1)tan^(-1)xdx-int_(0)^(1)tan^(-1)(x-1)dx`
`=int_(0)^(1)tan^(-1)xdx-int_(0)^(1)tan^(-1)(1-x-1)dx`
`=2int_(0)^(1)tan^(-1)xdx`
`=2[x tan^(-1)x-int(x)/(1+x^(2))dx]_(0)^(1)`
`=2[x tan^(-1)x-(1)/(2)log(1+x^(2))]_(0)^(1)`
`=2[(1tan^(-1)1-(1)/(2)log2)-(0-(1)/(2)log1)]`
`=(pi)/(2)-log2`


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