Question

Integrate the following function: $\frac{(1+log\ x)^2}{x}$

Answer 1

we have to integrate $\frac{(1+log\ x)^2}{x}$


Let 1 + logx = P


differentiating both sides,


0 + 1/x . dx = dP => dx = x. dP


now, $\int{\frac{(1+logx)^2}{x}}\,dx$


= $\int{\frac{P^2}{x}}\,x.dP$


= $\int{P^2}\,dP$


= $\left[\frac{P^3}{3}\right]+C$


put 1 + logx = P


so, $\int{\frac{(1+logx)^2}{x}}\,dx=\left[\frac{(1+logx)^3}{3}\right]+C$