## anonymous 5 years ago how do I solve int from e to 1 (2+lnx)^3 over x dx

1. anonymous

Set $u=(2+\log x)$Then$du=\frac{dx}{x}$and from the definition of our substitution,$e^u=e^{2+\log x}=xe^2 \rightarrow x=e^{u-2}$so the differential for du is$du = \frac{dx}{x}=\frac{dx}{e^{u-2}}\rightarrow dx=e^{u-2}du$Your integral becomes,$\int\limits_{e}^{1}\frac{(2+\log x)^3}{x}dx=\int\limits_{u_1}^{u_2}\frac{u^3}{e^{u-2}}e^{u-2}du=\int\limits_{u_1}^{u_2}u^3du=\frac{u^4}{4}|_{u_1}^{u_2}$

2. anonymous

So,$\int\limits_{e}^{1}\frac{(2+\log x)^3}{x}dx=\frac{(2+\log x)^4}{4}|_{e}^{1}=\frac{(2+0)^4}{4}-\frac{(2+1)^4}{4}$$=\frac{16-81}{4}=-\frac{65}{4}$