Can you add an extra $e^x$ when integrating?

Question

So I've been given this problem to solve (pretend it's a fraction or click the link to see the question please)

$$\int \frac{-26e^x-144}{e^{2x} + 13e^x + 36}dx$$

and I got this far:

$$-2\int\frac{13e^x + 72}{e^{2x} + 13e^x + 36}dx$$

The next step is a simple u-substitution which I am unable to do. I looked to a step-by-step calculator for help, and what I see is

$$\int\frac{13u+ 72}{u(u^2 + 13u + 36)}du$$

But I'm confused. Where did the extra $u$ come from?

I'm guessing that its like in the situation where you have $\int \cos(2θ)dθ$ and you have $u = 2θ$ and $du = 2dθ$, but there is no extra 2, so the end result is $\frac{1}{2}\int\cos(u)du$.

But I thought you could only do that with constants! You don't just add an extra $x^2$ when you need it. As thus, you shouldn't be able to add an extra $e^x$ just because you need it. Or am I missing something?

Answer 1

Let $f(z) = P(z)/Q(z)$ be a rational function in $z$ (that is to say, it is the quotient of two polynomials where $Q$ is not the zero polynomial). Then $$\int f(e^x) \, dx = \int \frac{f(e^x)}{e^x} \cdot e^x \, dx,$$ and with the substitution $u = e^x$, $du = e^x \, dx$, we obtain $$\int f(e^x) \, dx = \int \frac{f(u)}{u} \, du.$$ Since $f(u)$ is a rational function, so is $f(u)/u$. In your case, we have $$f(z) = \frac{-26z -144}{z^2 + 13z + 36},$$ hence $$\frac{f(u)}{u} = \frac{-26u - 144}{u(u^2 + 13u + 36)}.$$ After factoring a constant of $-2$ from the numerator, you get the result you mentioned.

Answer 2

I'm guessing that its like in the situation where you have $\int \cos(2θ)dθ$ and you have $u = 2θ$ and $du = 2dθ$, but there is no extra 2, so the end result is $\frac{1}{2}\int\cos(u)du$.

You're right, it is exactly like that. It's not the case that you can only do that with constants; in fact, you always have to do it. Any time you do a u-substitution, you always get an extra factor of $\frac{du}{dx}$.

To help understand why, you can think about it as though you're doing the substitution in reverse: instead of $u = e^x$, think about it as though it's $x = \ln u$. When you plug this into the integrand, you get $$\frac{13 e^x + 72}{e^{2x} + 13 e^x + 36} \quad\underset{x = \ln u}{\longrightarrow}\quad\frac{13 u + 72}{u^2 + 13u + 36}$$ It sounds like you have no trouble with that part. But you also have to make the same substitution in $dx$. That would get you $d(\ln u)$. (I'm overlooking a bunch of technical details that don't really matter in this case) But what does that mean? Well, if it were part of a derivative, I'm sure you could make sense of it: $$\frac{d(\ln u)}{du} = \frac{1}{u}$$ Or if it were part of a different derivative, I bet you could still make sense of it: $$\frac{d(\ln u)}{dx} = \text{use the chain rule} = \frac{d(\ln u)}{du}\frac{du}{dx} = \frac{1}{u}\frac{du}{dx}$$ Now, the $d(\ln u)$ from the integral isn't part of a derivative, but the math works basically the same way: $$d(\ln u) = \text{use the chain rule} = \frac{d(\ln u)}{du}du = \frac{1}{u}du$$ And that's where your extra factor of $\frac{1}{u}$ comes from. (Note: this is not a proof of that fact, just an illustrative argument that will hopefully help you make some sense of it.)

Answer 3

You almost have the right answer. Your second to last paragraph is the clue: you have to do this with all constants. The substitution here is $u=e^x$, i.e $du=e^xdx$, which means $\frac{du}{e^x}=dx$. Substituting, we get the extra $u$. The substitution in of the new $du$ is something we have to do for all substitutions, regardless of what the function is. Whether $du$ is expressed in terms of $dx$ with only constants, or polynomials, or whatever does not matter.

Answer 4

The extra $e^x$ comes from the $dx$. If you have $$\int \frac{13e^x+72}{e^{2x}+13e^x+36}dx$$ and you say $u=e^x$, then you also need to substitute the $dx$.

If $$u=e^x$$ then $$\frac{du}{dx}=e^x$$ $$du=e^xdx$$ $$\frac{du}{e^x}=dx$$ Since we said $u=e^x$, we can substitute that in here to get $$dx=\frac{du}{u}.$$ Then when you substitute these into your integral, you get $$\int \frac{13u+72}{u^2+13u+36}\cdot \frac{du}{u}$$ $$=\int \frac{13u+72}{u(u^2+13u+36)} du$$