How do you find the definite integral of #(lnx^3)/x# from #[1, e]#?

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Answer 1 · 2016-04-14T22:18:28

#3/2#

We want to find

#int_1^eln(x^3)/xdx#

First, note that the natural logarithm function can be simplified using the rule #ln(x^a)=aln(x)#. Thus, the integral is equivalent to

#=int_1^e(3ln(x))/xdx=3int_1^eln(x)/xdx#

Now, notice that this integral is ripe for substitution: we have present a function #ln(x)#, and its derivative, #1/x#. Thus, we let

#u=ln(x)" "=>" "(du)/dx=1/x" "=>" "du=1/xdx#

We can rearrange the integral to see our values of #u# and #du# more clearly:

#=3int_1^eln(x)1/xdx#

Now, before we substitute #u# and #du#, note that we will change the bounds of the integral by plugging the current bounds #[1,e]# into #u=ln(x)#:

#u(1)=ln(1)=0#
#u(e)=ln(e)=1#

Thus, the new bounds are #[0,1]# and the integral equals

#=3int_0^1ucolor(white).du#

To find this integral, recall that #u=u^1# and use the rule:

#intu^ndu=u^(n+1)/(n+1)+C#

Hence,

#intu^1du=u^2/2+C#

So, evaluating this on #[0,1]#, we obtain

#=3(u^2/2)]_0^1=3[(1^2/2)-(0^2/2)]=3(1/2-0)=3/2#