Integration of the Natural Logarithm Raised to a Power
Formulas
$$ \int {{(\ln x)}^{ n }} dx=x{{(\ln x)}^{ n }}-\int n{{(\ln x)}^{ n-1 }}dx $$
Explanation
When solving integral problems, it’s not uncommon to encounter this type of problem. Solving these problems directly through integration by parts can be extremely time-consuming. First, let’s try to find a rule. Given $f(n)=\int {{(\ln x)}^{ n }} dx$ (where $n=1,2,3…$),
$$ \begin{align*} f(1) =& x(\ln|x|-1)+C \\ f(2) =& x{(\ln|x|)^{ 2 }-2\ln|x|+2}+C \\ f(3) =& x{(\ln|x|)^{ 3 }-3(\ln|x|)^{ 2 }+6\ln|x|-6}+C \\ f(4) =& x{(\ln|x|)^{ 4 }-4(\ln|x|)^{ 3 }+12(\ln|x|)^{ 2 }-24\ln|x|+24}+C \end{align*} $$
Looking at the result of integrating up to the fourth power, a certain rule becomes apparent. Except for $f(n)$ in $x(\ln|x|)^{ n } $, it is multiplying by $-n$ after removing the integration constant from $f(n-1)$.
$$ f(1)=x(\ln|x|-1)+C $$
Multiply $\downarrow \ln|x|-1$ by $-2$.
$$ f(2)=x{(\ln|x|)^{ 2 }-2\ln|x|+2}+C $$
Multiply $\downarrow –2\ln|x|+2$ by $-3$.
$$ f(3)=x{(\ln|x|)^{ 3 }-3(\ln|x|)^{ 2 }+6\ln|x|-6}+C $$
Multiply $\downarrow -3(\ln|x|)^{ 2 }+6\ln|x|-6$ by $-4$.
$$ f(4)=x{(\ln|x|)^{ 4 }-4(\ln|x|)^{ 3 }+12(\ln|x|)^{ 2 }-24\ln|x|+24}+C $$
If we generalize the above process for $n$,
$$ \int {{(\ln x)}^{ n }} dx=x{{(\ln x)}^{ n }}-\int n{{(\ln x)}^{ n-1 }}dx $$
As you can see from the formula, it’s challenging to write down the sequence just by looking at the degrees, so it’s a good idea to memorize the formula up to the second or third degree.
Reference
The natural logarithm and the exponentials of e are closely related, hence they often appear in similar forms.
$$ \int x^{ n }e^{ x }dx=x^{ n }e^{ x }-\int nx^{ n-1 }e^{ x }dx $$
The only difference from the introduced degrees is the degree of $x$.
$$ \int { \lambda x { e }^{ \lambda x }dx }=\left( x-\frac { 1 }{ \lambda } \right) {e}^{ \lambda x }+c $$
$n=1$, and cases where $x$ is a constant multiple, are used more often than one might think. Even if the shape isn’t exactly like that, applying the formula after an appropriate transformation can make the solution much simpler.