How do you integrate #int x^3 ln x^2 dx # using integration by parts?

Question

How do you integrate #int  x^3   ln x^2     dx  # using integration by parts?

Emily Ammerman · Accepted Answer

#x^4/8(4lnx-1)+C#

Let #I=intx^3lnx^2dx#

Using the Subst. #x^2=t", we have, "2xdx=dt, or, xdx=1/2dt.# Hence,

#I=int(x^2lnx^2)xdx=1/2inttlntdt.#

Now, to this end, we will use the Rule of I ntegration byParts arts :

#IBP : intuvdt=uintvdt-int{(du)/dtintvdt}dt.#

We take #u=lnt rArr (du)/dt=1/t, &, v=trArr intvdt=t^2/2#

#:. I=1/2[t^2/2lnt-int(1/t*t^2/2)dt]#

#=t^2/4lnt-1/4inttdt=t^2/4lnt-t^2/8#

Returning #t" by "x^2,# we have,

#I=x^4/4ln(x^2)-1/8x^4#

Here, we use the Rule of Logarithm # : lnx^2=2lnx# & get,

#I=x^4/2lnx-x^4/8=x^4/8(4lnx-1)#

Samantha Adkison · Answer

undefined To integrate ∫x^3 ln(x^2) dx using integration by parts, follow these steps:

1. Choose u and dv:
   Let u = ln(x^2) and dv = x^3 dx.

2. Find du and v:
   Differentiate u to find du, and integrate dv to find v.

du = (1/x^2) * 2x dx
      = 2/x dx

v = ∫x^3 dx
     = (1/4)x^4

3. Apply the integration by parts formula:
   ∫u dv = uv - ∫v du

∫x^3 ln(x^2) dx = (ln(x^2)) * (1/4)x^4 - ∫(1/4)x^4 * (2/x) dx

4. Simplify and integrate the remaining integral:
   ∫(1/4)x^4 * (2/x) dx = (1/2) ∫x^3 dx
                         = (1/2) * (1/4)x^4
                         = (1/8)x^4

5. Combine the results:
   ∫x^3 ln(x^2) dx = (1/4)x^4 ln(x^2) - (1/8)x^4 + C

Therefore, the integral of x^3 ln(x^2) dx using integration by parts is (1/4)x^4 ln(x^2) - (1/8)x^4 + C, where C is the constant of integration.