How do you evaluate the definite integral #int (1 + ln x)^2 / x# from 1 to e?
2 Answers
Mar 16, 2016
We can see that
Note
Mar 26, 2016
Explanation:
Let
This implies that
Thus, we have
#int_1^e(1+lnx)^2/xdx=int_0^1(1+u)^2du#
Notice that the bounds also changed, since we substituted in
#u(1)=ln(1)=0#
#u(e)=ln(e)=1#
Expand
#=int_0^1(1+2u+u^2)du#
We can integrate
#=[u+u^2+u^3/3]_0^1=(1+1^2+1^3/3)-(0+0^2+0^3/3)#
#=7/3#