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How to integrate $$ \int_{1}^{e} (x+1)e^{x}\ln{x}dx$$

I used following ways:

  • integration by parts

I first split the function into $$ \int_{1}^{e} (x)e^{x}\ln{x}dx + \int_{1}^{e} e^{x}\ln{x}dx$$

And then let $$ I_1 = \int_{1}^{e} (x)e^{x}\ln{x}dx $$ and $$I_2 = \int_{1}^{e}e^{x}\ln{x}dx$$ And I get $$I_2 = e^{x}(x-1)$$ But while solving for $I_1$ I stick to $$\int_{1}^{e}\frac{e^{x}}{x}dx$$ - substitutions I used $$ lnx = t $$ $$x=e^{t}$$ $$dx =e^{t}dt$$

But here the integral will become

$$ \int_{0}^{1}(e^{t}+1)e^{e^{t}}tdt$$
What to do next?

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Noting $$ d(xe^x)=(x+1)e^x$$ one has \begin{eqnarray} \int_1^e(x+1)e^xdx\ln x&=&\int_1^e\ln xd(xe^x)\\ &=&xe^x\ln x\big|_1^e-\int_1^e xe^xd\ln x\\ &=&xe^x\ln x\big|_1^e-\int_1^e e^xdx \end{eqnarray} and the rest is easy.

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$$I=\int_1^e(x+1)e^x\ln(x)dx$$ $\frac{dv}{dx}=(x+1)e^x$ so $v=xe^x$ and $u=\ln(x)$ so $\frac{du}{dx}=\frac{1}{x}$ $$I=\left[xe^x\ln(x)\right]_1^e-\int_1^ee^xdx=\left[e^x\left(x\ln(x)-1\right)\right]_1^e=e^{e+1}$$

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