This example provides us with a better understanding of how "Integration by parts" works. Building an extensive knowledge of Integral calculus requires a lot of work, but once the basics are fully understood, it follows naturally that one would go into a more deep approach.
Recall: ∫fg′=fg−∫f′g (Integration by parts), where f and g are well-defined functions
Approach:
1) Think before you act
2) Think two steps ahead
3) Be clear and not verbose
Solution:
∫xlnx dx=
= ∫ [(x^2)/2]' lnx dx=
= [(x^2)/2]lnx - ∫[(x^2)/2] (lnx)' dx=
= [(x^2)/2]lnx - 1/2∫(x^2)*(1/x) dx=
= [(x^2)/2]lnx - 1/2∫xdx=
= [(x^2)/2]lnx - 1/2 * (x^2)/2 +C (don't forget the constant)=
= (simple manipulation) [(x^2)(2lnx -1)]/4 +C
Now remember, Maths can be very easy once you take action. The best car in the world will not take you to the right place if you don't know where you want to go. #naturalenthusiasm