Question

calculate area under (ln(x))²/x, between x=1, x=e

Answer 1

\[\Large A=\int_1^e \frac{(\ln(x))^2}{x}dx \]

Answer 2

use an appropriate u-substitution. can you think of one?

Answer 3

you can use the integration-by-part formula, twice.
The first time, take \(f(x)=(\ln(x))^2\) and \(g'(x)=\frac 1x\). Then
\[A=[\ln(x)^2\ln(x)]^e_1 - \int_1^e (2\ln(x)\frac 1x)(-\frac 1{x^2})dx
\]
put again the \(\ln(x)\) in \(f\) and \(g'\) is the rest.

Answer 4

I believe \[\Large u=\ln x \]
such that:
\[\Large \frac{du}{dx}=\frac{1}{x} \]
might be a bit easier for this kind of integral:

\[\Large \int_a^b (\ln x)^2 \frac{1}{x}dx= \int_a^bu^2du \]

Answer 5

"the rest": I rather mean: \(g'\) is the remaining function.