`int(xlnx)/((x^2-1)^(3/2))dx`

int(x)/((1+x)^((3)/(2)))dx=

Evaluate : (i) int(3x^(2))/((1+x^(6)))dx (ii) int(x^(3))/((x^(2)+1)^(3))dx (iii) (x^(8))/((1-x^(3))^(1//3))dx

int_2^3 (dx)/(x^2-1)

Evaluate: (i) int(x^(2)+x+5)/(3x+2)dx (ii) int(2x+3)/((x-1)^(2))dx

The value of int_(0)^(oo)(xlnx)/((1+x^(2))^(2))dx is a) oo b) 1//2 c)2 d)0

int(2x^(2)+3)sqrt(x+2)dx( ii) int(x^(2)+3x+1)/((x+1)^(2))dx

int(3x^(2)+1)/((x^(2)-1)^(3))dx

int(x^(2)+3x-1)/((x+1)^(2))dx

int(3x-2)/((x+1)^(2)(x+3))dx