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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Set \[u=(2+\log x)\]Then\[du=\frac{dx}{x}\]and from the definition of our substitution,\[e^u=e^{2+\log x}=xe^2 \rightarrow x=e^{u2}\]so the differential for du is\[du = \frac{dx}{x}=\frac{dx}{e^{u2}}\rightarrow dx=e^{u2}du\]Your integral becomes,\[\int\limits_{e}^{1}\frac{(2+\log x)^3}{x}dx=\int\limits_{u_1}^{u_2}\frac{u^3}{e^{u2}}e^{u2}du=\int\limits_{u_1}^{u_2}u^3du=\frac{u^4}{4}_{u_1}^{u_2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So,\[\int\limits_{e}^{1}\frac{(2+\log x)^3}{x}dx=\frac{(2+\log x)^4}{4}_{e}^{1}=\frac{(2+0)^4}{4}\frac{(2+1)^4}{4}\]\[=\frac{1681}{4}=\frac{65}{4}\]
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