`int(2x+3)/((x-1)(x^2+1))dx =log_e{(x-1)^(5/2)(x^2+1)^a-1/2 tan^-1 x+C,x > 1` where `C` is any arbitrary constant, then the value of '`a'` is

For real numbers alpha, beta, gamma and delta , if int((x^(2)-1)+tan^(-1)((x^(2)+1)/(x)))/((x^(4)+3x^(2)+1)tan^(-1)((x^(2)+1)/(x)))dx = alpha log_(e)(tan^(-1)((x^(2)+1)/(x)))+beta "tan"^(-1)((gamma(x^(2)-1))/(x))+delta tan ((x^(2)+1)/(x))+C where is an arbitrary constant, then the value of 10(alpha+betagamma+delta) is equal to........

int(sin^(-1)x)/(sqrt(1-x^(2)))dx is equal to Where, C is an arbitrary constant.

If int\ x ln(1+1/x)\ dx = p(x)\ ln(1+1/x)+1/2 x - 1/2 ln(1+x) + c, c being arbitarary constant, then

If int(3x+2)/(2x^(2)+2x+1)dx = m log(2x^(2)+2x+1)+(1)/(2)tan^(-1)u+c , then

If int (2x^(2)+3)/((x^(2)-1)(x^(2)+4)) dx = a log((x-1)/(x+1)) +b tan^(-1) (x/2) +c then the values of a and b are

The integral int(2x^12+5x^9)/((x^5+x^3+1)^3)dx is equal to: where C is an arbitrary constant.

If int1/((x^(2)-1))log((x-1)/(x+1))dx=A[log((x-1)/(x+1))]^(2)+c , then A =

If int1/((x^(2)-1))log((x-1)/(x+1))dx=A[log((x-1)/(x+1))]^(2)+c , then A =