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

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Answer 1 · 2016-10-05T17:47:01

#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)#