How do you integrate #int x^3 ln x^4 dx # using integration by parts?

First, let's rewrite #ln(x^4)# as #4ln(x),# as #ln(x^a)=aln(x)#

#intx^3ln(x^4)dx=int4x^3ln(x)dx=4intx^3ln(x)dx# (We can factor out #4,# it's just a constant.)

Make the following choices for #u# and #dv:#

#u=ln(x)#

#dv=x^3dx#

Thus, #du# and #v# become:

#du=1/xdx#

#dv=intx^3dx=1/4x^4# (We're not going to include the constant just yet. It can wait until the end.)

Now it becomes apparent why we made the choices we did for #u# and #v.# If we chose #dv=ln(x),# we would have to integrate #ln(x)# for #v,# and would run into our original problem.

Plug in relevant values:

#uv-intvdu=1/4x^4ln(x)-int1/4x^4/xdx#

We can simplify the integral here:

#1/4x^4ln(x)-int1/4x^4/xdx=1/4x^4ln(x)-1/4intx^(cancel(4)3)/cancelx#

Thus, we have

#1/4x^4ln(x)-1/4intx^3dx=1/4x^4ln(x)-(1/4)(1/4)x^4+C=1/4x^4ln(x)-1/16x^4+C#

Let's not forget the #4# we originally factored out...

#4(1/4x^4ln(x)-1/16x^4+C)=x^4ln(x)-1/4x^4+C#

#C# remains unchanged; a constant multiplied by a constant remains a constant.

So,

#intx^3ln(x^4)dx=x^4ln(x)-1/4x^4+C#

Here's another approach:

Consider substituting first #w=x^4# and #("d"w)/("d"x)=4x^3# into the integral #int\ x^3ln(x^4)\ "d"x#.

This gives #1/4int\ ln(w)\ "d"w#.

Now use integration by parts #int\ u\ "d"v=uv-int\ v\ "d"u#. Use #u=ln(w)# and #"d"v="d"w#. Then find that #"d"u=("d"w)/w# and #v=w#.

Now, we have #1/4int\ ln(w)\ "d"w=1/4(wln(w)-int\ "d"w)#
or simply #1/4wln(w)-1/4w+C#.

Substitute back #w=x^4#.

Our final answer is thus #1/4x^4ln(x^4)-1/4x^4+C=x^4ln(x)-1/4x^4+C#.