This my big problem in my integration problem, $\displaystyle \int{e^x \ln x$.In this case i choose u=$\displaystyle \ln x$ and $\displaystyle \frac{de^x} {dx}=e^x$.Then i stuck on this $\displaystyle \int{\frac{e^x}{x}}$.please give me a solution
This my big problem in my integration problem, $\displaystyle \int{e^x \ln x$.In this case i choose u=$\displaystyle \ln x$ and $\displaystyle \frac{de^x} {dx}=e^x$.Then i stuck on this $\displaystyle \int{\frac{e^x}{x}}$.please give me a solution
I agree with your method of integration by parts, so you should end up with $\displaystyle \displaystyle \begin{align*} \int{e^x\ln{(x)}\,dx} = e^x\ln{(x)} - \int{\frac{e^x}{x}\,dx} \end{align*}$. The second function does not have a solution in terms of the elementary functions, but Wolfram tells us that this is the Exponential Integral.
Of course, a series solution could be appropriate...
$\displaystyle \displaystyle \begin{align*} e^x &= \sum_{n = 0}^{\infty}{\frac{x^n}{n!}} \\ \frac{e^x}{x} &= \sum_{n = 0}^{\infty}{\frac{x^{n-1}}{n!}} \\ &= \frac{1}{x} + \sum_{n = 1}^{\infty}{\frac{x^{n-1}}{n!}} \\ \int{\frac{e^x}{x}\,dx} &= \int{\frac{1}{x} + \sum_{n=1}^{\infty}{\frac{x^{n-1}}{n!}}\,dx} \\ &= \ln{|x|} + \sum_{n = 1}^{\infty}{\frac{x^n}{n \cdot n!}} + C \end{align*}$