This integration technique should never be forgot!

[Lorena_Santoro]

https://youtu.be/virrFtQKWy0

[Guest]

I really dislike having to open a link, especially to "you-tube" to see what should be posted on this site! In any case, the integral of ln(x) is covered in any "Calculus II" course.

I will agree that it is a clever trick!

To integrate [tex]\int ln(x)dx[/tex] use "integration by parts" (even though there is only one function!).

The "integration by parts" formula is [tex]\int udv= uv- \int vdu[/tex]. It is the inverse of the "product rule" for differentiation: write d(uv)= udv+ vdu as udv= d(uv)- vdu and integrate both sides, [tex]\int udv= uv- \int vdu[/tex].

Here, although there is only one function, let u= ln(x) and dv= dx. Then [tex]du= \frac{1}{x}dx[/tex] and [tex]v= x[/tex] so [tex]\int ln(x)dx[/tex][tex]= \left(ln(x)\right)\left(x\right)-[/tex][tex]\int x\left(\frac{1}{x}dx\right)=[/tex][tex]xln(x)- \int dx= xln(x)- x[/tex].

[Guest]

Great answer!

[Lorena_Santoro]

Correct!