# Evaluate the following definite integral. \int_{1}^{e}\ln 2 xdx

Evaluate the following definite integral.
${\int }_{1}^{e}\mathrm{ln}2xdx$
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Clelioo
${\int }_{1}^{e}\mathrm{ln}2xdx$
Substitute u =2x
$\frac{du}{dx}=2$
$dx=\frac{1}{2}du$
$=\frac{1}{2}{\int }_{2}^{2e}\mathrm{ln}\left(u\right)du$
Now solve ${\int }_{2}^{2e}\mathrm{ln}udu$
Integrate by parts:
$\int f{g}^{\prime }=fg-\int {f}^{\prime }g$
$f=\mathrm{ln}u,{f}^{\prime }=\frac{1}{u}$
g'=1, g=u
$⇒{\left[u\mathrm{ln}\left(u\right)\right]}_{2}^{2e}-{\int }_{2}^{2e}1du$
Now solve ${\int }_{2}^{2e}1du={\left[u\right]}_{2}^{2e}$
Plug in the solved integrals:
${\left[u\mathrm{ln}\left(u\right)\right]}_{2}^{2e}-{\int }_{2}^{2e}1du$
$={\left[u\mathrm{ln}\left(u\right)-u\right]}_{2}^{2e}$
Plug in the solved integrals:
$\frac{1}{2}{\int }_{2}^{2e}\mathrm{ln}\left(u\right)du$
$={\left[\frac{u\mathrm{ln}\left(u\right)}{2}-\frac{u}{2}\right]}_{2}^{2e}$
Undo the substitution u=2x
$={\left[x\mathrm{ln}\left(2x\right)-x\right]}_{1}^{e}$
$=\left[e\mathrm{ln}\left(2e\right)-e\right]-\left[1\mathrm{ln}\left(2\right)-1\right]$
$=e\mathrm{ln}\left(2e\right)-e-\mathrm{ln}\left(2\right)+1$
Jeffrey Jordon