How do you integrate #int 4 e^x ln x dx # using integration by parts?

1 Answer
Apr 4, 2016

You can try, but you'll find that this integral can produce an infinity of integrals if you keep applying integration by parts over and over.

Explanation:

Integration by parts is:
#intudv=uv-intvdu#
A helpful acronym for choosing the #u# of IbP is LIATE :
L - logarithmic (ex. #lnx#, #log_3x#)
I - inverse trigonometric (ex. #tan^(-1)x, arcsin(x)#)
A - algebraic (ex. #x^2#, #1/x#)
T - trigonometric (ex. #sinx#, #cosx#)
E - exponential (ex. #e^x#, #2^x#)

In our case, we have a logarithmic (#lnx#) and exponential (#4e^x#) function. In our LIATE list, logarithmic functions come first, so our #u# will be #lnx#. That means everything else, namely #4e^xdx# will be #dv#:
#u=lnx->(du)/dx=1/x->du=1/xdx#
#dv=4e^xdx->intdv=int4e^xdx->v=4e^x#

Applying the parts formula:
#int4e^xlnxdx=(lnx)(4e^x)-int4e^x1/xdx#
#color(white)(XX)=4e^xlnx-4inte^x1/xdx#

We have to do integration by parts a second time for the new integral #4inte^x1/xdx#:
#u=1/x->(du)/dx=-1/x^2->du=-1/x^2dx#
#dv=e^xdx->intdv=inte^xdx->v=e^x#

Applying what we found:
#int4e^xlnxdx=4e^xlnx-4(1/xe^x-inte^x(-1/x^2dx))#
#color(white)(XX)=4e^xlnx-4(e^x/x+inte^x1/x^2dx)#

Ahhhhh! We have to use another integration by parts! But before you do, notice what's going on. Each time we integrate by parts, we end up with something like #e^x1/x# in a new integral. This is, in fact, a pattern. If we keep integrating by parts, we'll get one integral after another, in the form #int(-1)^n(e^x/x^n)dx# - and it will never stop. So this integral continues on forever with no solution.