Question

how do I solve int from e to 1 (2+lnx)^3 over x dx

Answer 1

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,\[\int\limits_{e}^{1}\frac{(2+\log x)^3}{x}dx=\int\limits_{u_1}^{u_2}\frac{u^3}{e^{u-2}}e^{u-2}du=\int\limits_{u_1}^{u_2}u^3du=\frac{u^4}{4}|_{u_1}^{u_2}\]

Answer 2

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