How do you find the definite integral of #(lnx^3)/x# from #[1, e]#?
1 Answer
Explanation:
We want to find
#int_1^eln(x^3)/xdx#
First, note that the natural logarithm function can be simplified using the rule
#=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
#u=ln(x)" "=>" "(du)/dx=1/x" "=>" "du=1/xdx#
We can rearrange the integral to see our values of
#=3int_1^eln(x)1/xdx#
Now, before we substitute
#u(1)=ln(1)=0#
#u(e)=ln(e)=1#
Thus, the new bounds are
#=3int_0^1ucolor(white).du#
To find this integral, recall that
#intu^ndu=u^(n+1)/(n+1)+C#
Hence,
#intu^1du=u^2/2+C#
So, evaluating this on
#=3(u^2/2)]_0^1=3[(1^2/2)-(0^2/2)]=3(1/2-0)=3/2#