Integrate of e^x * lnx dx?

Question

∫ e^x * lnx dx

Answer 1

Using integration by parts with u = ln x, dv = e^x dx, this equals

e^x ln x - integral (e^x/x) dx.



However, this last integral is known to be impossible to integrate in terms of (finite combinations of) elementary functions. So, this integral is impossible to integrate in terms of elementary functions, too.

Answer 2

Integration by making use of areas isn't required for this concern. Integration by making use of substitution is defined with a link in the source. combine the unique integrand by making use of substitution: ? cos(lnx) / x dx permit u = lnx, du / dx = a million / x du = dx / x ? cosu du sinu + C on account that u = lnx, sin(lnx) + C

Answer 3

a bit late, but whatever. if i m right (and correct me if i m not) then the previous two answers are incorrect.



note that x is unequal to 0. integrate by parts, setting u=e^x and dv=lnxdx

then du=e^xdx and v=xlnx-x



thus, we have integral(e^x*lnx)dx = e^x(xlnx-x) - integral(e^x(xlnx-x)dx

= e^x(xlnx-x) + integral(xe^x)dx - integral(xe^x(lnx))dx (*)



knowing that integral(xe^x)dx = xe^x - e^x + c, and then integrating the second integral on the LHS of (*) by parts, setting u=x and dv=e^xlnxdx, (*) becomes



e^x(xlnx-x)+xe^x-e^x-x*[integral(e^xlnx)dx]+integral(e^xlnx)dx (=integral(e^xlnx)dx)



cancelling integral(e^xlnx)dx from both sides of this equation, and then rearranging to make

x*[integral(e^xlnx)dx] the subject, we now have:



x*[integral(e^xlnx)dx] = e^x(xlnx-x)+xe^x-e^x



giving the result

integral(e^xlnx)dx = e^x(lnx-1)+e^x-[(e^x)/x]+C (allowed since x is unequal to 0)

Answer 4

google "integral calculator"



the company that runs the site is called wolfram i believe. Its been very useful to me.



If you need to show your work then you need to use integration by parts.



let dv = e^x and u = ln[x] .... then du = 1/x and v = e^x



look up the integration by parts equation on google if you don't already know it. From what I've given you you should have a simple integral left to solve for. if you have more problems IM me on AIM at run4it987