`int(logx-1)/((logx)^2)dx `

underset2oversete int (1/logx-1/(logx)^2)dx

Evaluate the following integrals: int[(1)/(logx)-(1)/((logx)^2)]dx

Evalaute: int((logx-1)/(1+(logx)^2))^2dx

The value of int{(logx-1)/(1+(logx)^(2))}^(2)dx is equal to -

int{(logx-1)/(1+(logx)^(2))}^(2) dx is equal to

Evalaute: int((logx-1)/(1+(logx)^2))^2dx

int{(logx-1)/(1+(logx)^(2))}^(2) dx is equal to

int{(logx-1)/(1+(logx)^(2))}^(2)dx is equal to a) (logx)/((logx)^(2)+1)+c b) (x)/(x^(2)+1)+c c) (xe^(x))/(1+x^(2))+c d) (x)/((logx)^(2)+1)+c

If int[(logx-1)/(1+(logx)^2)]^2dx=f(x)/(1+(g(x))^2)+c , then (A) f(x)=x (B) f(x)=x^2 (C) g(x)=logx (D) g(x)=(logx)^2