Am I allowed to break up $\ln^3x$ as $\ln x^2\ln x$ in order to solve a Integral (by parts)?

Question

When I solve $\int \sec x^3\,dx$ by parts I can do $u=\sec x$, $du=\sec x\tan x\,dx$, $v=\tan x$ and $dv=\sec^2x$ and solve this way: $$\int \sec^2x\sec x=\sec x\tan x-\int \tan^2x\sec \tan x$$ When I take these I get the right answer. However, if I try something similar for $\ln^3x$ it does not work. I only get the right answer if I choose $u=\ln^3x$ and $v=x$. What am I missing?

Answer 1

Let $I$ be the indefinite integral

$$I=\int \log^3(x)\,dx$$

We can indeed integrate by parts using $u=\log^2(x)$ and $v=x\log(x)-x$. Proceeding, we find

$$\begin{align} I&=x\log^3(x)-x\log^2(x)-2\int (\log^2(x)-\log(x))\,dx\\\\ &=x\log^3(x)-x\log^2(x)+2x\log(x)-2x-2\int \log^2(x)\,dx \tag 1 \end{align}$$

We continue by integrating by parts the integral on the right-hand side of $(1)$, this time using $u=\log^2(x)$ and $v=x$. We find that

$$\begin{align} \int \log^2(x)\,dx&=x\log^2(x)-2\int \log(x)\,dx\\\\ &= x\log^2(x)-2x\log(x)+2x \tag 2 \end{align}$$

Substituting $(2)$ into $(1)$ yields

$$\bbox[5px,border:2px solid #C0A000]{I=x\log^3(x)-3x\log^2(x)+6x\log(x)-6x+C}$$

Answer 2

Integrating by parts, $$I_n=\int(\ln x)^n\ dx=(\ln x)^n\int\ dx-\int\left(\dfrac{d(\ln x)^n}{dx}\int dx\right)dx$$

$$=x(\ln x)^n-n\int(\ln x)^{n-1}dx$$

$$\implies I_n=x(\ln x)^n-nI_{n-1}$$