Why can I use u-substitution in $\int \frac{x^2}{x^3-7}dx$? Here is the problem from the textbook:
$$\int \frac{x^2}{x^3-7}dx$$
I don't understand why u substitution works on this problem, as in the explanation from the textbook I can only use it when I have two factors where one of them being the derivative of the otherone.
In the solution the enumerator is being treated as $\operatorname{g'}(x)$ and the first step of the solution looks like this:
$$\frac{1}{3}\int \frac{3x^2}{x^3-7}dx$$
Can we just add the factor 3 to x² because we're multiplying the whole integral by 1/3 which neutralizes the factor of 3 again?
 A: You can do a $u$-substitution when you have something containing a function, and that something is multiplied by the derivative of that function. In other words, if you have $∫(())′()$, you can use a u substitution.
In your case you have $f(x)=\frac{1}{x-7}$ and $g(x)=x^3$ to give $f(g(x))g'(x)=\frac{1}{x^3-7}\cdot3x^2=\frac{3x^2}{x^3-7}.$ So to get rid of the extra factor of $3$, you multiply it by $\frac{1}{3}$. Then you substitute $u=g(x)=x^3.$
A: Yes, that is correct.
Since we take $g(x)=x^3-7$, and since $g'(x)=3x^2$, we would like to have $3x^2$ in the numerator. So, we do\begin{align}\int\frac{x^2}{x^3-7}\,\mathrm dx&=\frac13\int\frac{3x^2}{x^3-7}\,\mathrm dx\\&=\frac13\int\frac{g'(x)}{g(x)}\,\mathrm dx\end{align}
A: To add to this discussion, although $u$-substitution has some technical conditions it should meet; they're extremely lenient!
Consider the somewhat-ridiculous substitution $u=\ln(x)$ in
$$\int \ln(x)dx$$
$u=\ln(x)\implies du=\frac{dx}{x}\implies dx=e^udu$
$$\implies\int \ln(x)dx=\int ue^udu$$
Now, integration by parts is trivial
$$\implies\int ue^udu=e^u(u-1)+C$$
$$=x(\ln(x)-1)+C$$
I find this to be a very fun integral. Especially because it doesn't easily seem to be of the form $∫(())′()$
A: your integral is of a nice form which is:
$$\int\frac{f'(x)}{f(x)}dx=\int\frac{df}{f}=\ln|f|+C$$
and in general integrals of the form:
$$\int f'(x)g(f(x))dx=\int g(u)du=g^{(-1)}(f(x))+C$$
