`int_0^1xtan^(- 1) x dx=`

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int_0^1xtan^(-1)xdx

Evaluate: int_0^1xtan^(-1)xdx

int_0^(1) tan^(-1) x dx =

int_(0)^(1)xtan^(-1)xdx

int_(0)^(1)xtan^(-1)xdx=

If 2int_0^1 tan^(-1)x dx= int_0^1 cot^(-1)(1-x+x^2) dx then int_0^1 tan^(-1)(1-x+x^2) dx=

I : int_(0)^(pi//2)(2tan""(x)/(2)+xsec^(2)""(x)/(2))dx=pi II : int_(0)^(1)xTan^(-1).xdx=(pi)/(4)-(1)/(2)

The value of int_(-a)^a(cos^(- 1)x-sin^(- 1)sqrt(1-x^2))dx is (a>0) there int_0^acos^(- 1)x dx=A) is

The value of int_(-a)^a(cos^(- 1)x-sin^(- 1)sqrt(1-x^2))dx is (a>0) where (int_0^acos^(- 1)x dx=A) is