How do you factor $\ln(x)$? In the following equation how come $\ln(x)$ is part of the factor? I would understand  if $\ln(x^2)$. But how can you include $\ln(x)$ into the factoring? If you expanded this would $\ln(x)$ be unaffected from $x^2$?
$$x^2 + 3x^2 \ln(x) = x^2 ( 1 + 3\ln(x))$$
 A: Consider $\ln(x)$ as its own object. When you pull a factor of $x^2$ out of each term, the first term leaves a $1$ and the second term a $3\ln(x)$. Alternatively, going in the other direction, when you distribute the $x^2$ to both terms the second is $x^2\times2\ln(x)$ which has nothing to do with $\ln(x^2)$.
A: I'm really not sure where you're getting $\ln (x^2)$ from, but here's an attempt to clear this up:
Suppose that we just said $Y = \ln(x)$.  Then we could write this as
$$
x^2 + 3x^2 \ln(x) = x^2 + 3x^2 Y
$$
Now, $Y$ is just a number.  All we need to do is treat it as such, and factor the expression.  So, we have
$$
x^2 + 3x^2 Y = x^2 + x^2\cdot (3Y) = x^2(1 + 3Y) = x^2(1 + 3\ln(x))
$$
A: Have you ever factored an expression using a substitution (e.g. $2a^6-a^3-6$). You can factor $x^2+3x^2\ln(x)$ using a substitution as well. Let $\ln(x)=a$. Then $x^2+3x^2\ln(x)=x^2+3x^2a$. Factoring gives us $x^2(1+3a)$. Substitute $\ln(x)$ back into $a$. The answer is:
$$\boxed{x^2+3x^2\ln(x)=x^2\left[1+3\ln(x)\right]}$$
