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?


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.


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.