5.3
Evaluate $\int \frac{x^3}{9-x^4}dx$ using u substitution and the log identity for integrating $x^{-1}$
Can somebody verify this solution for me? Thanks!!
Let $u=9-x^4$. Then $\frac{du}{dx}=-4x^3$ and so $\frac{du}{-4x^3}=dx$. Thus we have:
$=\int \frac{x^3}{9-x^4}dx$
$=\int \frac{x^3}{u} \frac{du}{-4x^3}$
$=\frac{1}{-4} \int \frac{1}{u}du$
$=\frac{1}{-4} ln|u| + C$
$=\frac{1}{-4} ln|9-x^4| +C$
Okay, I just want to talk about something that I've been debating saying. Let's go back to this step:
$=\frac{1}{-4} \int \frac{1}{u}du$
Now, go look back and what I said the next equality sign was. Looks fine, right? But shouldn't it technically be:
$=\frac{1}{-4} (ln|u|+C)$
$=\frac{1}{-4} ln|u| - \frac{C}{4}$
See the difference? It's very subtle, but when you do it this way, which looks more correct, the constant of integration $C$ ends up with a $\frac{-1}{4}$ attached to it... It makes sense, since $\int \frac{1}{u} du = ln|u|+C$.
The point here is that the variable $C$ is called an arbitrary constant of integration for a reason. The point is we have no clue what it is, so if we multiply it by $\frac{-1}{4}$, we might as well just call it $C$ still because it can be anything anyway.... Note that this is not the case when we are doing a problem that gives "initial conditions", those problems where we solve for $C$ or plug something in for $C$, because then $C$ is no longer arbitrary.
Anyway, if your confused, that's okay, just kind of wanted to say this.
You can do it much easier: the chain rule and basic integration tells us that if we have a function $\;f(x)\;$ such that $\;\int f(x)\,dx= F(x)\;$ (meaning: $\;F(x)\;$ is a primitive of $\;f(x)\;$ , then for any differentiable funcion $\;g(x)\;$ we have that $\;\int g' (x) f(g(x))\,dx=F(g(x))\;$ .This, of course, is just the principle for the substitution method, but we can make it shorter if we recognize the pattern.
In our case, observe that $\;(9-x^4)'=-4x^3\;$, so that we can write
$$\int\frac{x^3}{9-x^4}dx=-\frac14\int\frac{\overbrace{-4x^3}^{=(9-x^4)'}}{9-x^4}dx=-\frac14\log|9-x^4|+C (=\text{a constant})$$