4

I was reviewing one of my integral problems, and I ran into this problem:

$$\int \ln(2x+4)dx $$

I did integration by parts and got this:

$$\ln(2x+4)x - \int \frac{2x}{2x+4}du $$

For the second integral, I did a u-sub:

$$ \frac{1}{2} \int \frac{u-4}{u} dx $$ Where $u=2x+4$, but the issue here is that my review notes told me that I can't do a sub here. The review notes told me to do a polynomial division instead. If I do it my way, I get the answer wrong but there was no explanation of why I can't do a sub for that type of integral.

mohhamed rafi
  • 45
  • 2
    Are you sure that your teacher said that you can not do a substitution. I see no reason that doing one would cause a mistake. Perhaps you made a separate error when doing the substitution. Anyway, there is no reason to do one, since $;\int\frac{2x}{2x+4}dx = \int\frac{2x +4}{2x+4}dx - \int\frac{4}{2x+4}dx,;$ which is easily integrated. Perhaps the idea is that substitution is something of a last resort (and in this case unnecessary complication). – user2661923 Mar 05 '20 at 00:22
  • 2
    You must convert to $du$ and not merely leave the $dx$. – GEdgar Mar 05 '20 at 00:23
  • @GEdgar Oh sorry that was a mistake when I typed that, I ment to write $du$. – mohhamed rafi Mar 05 '20 at 00:25
  • 1
    You can do u-sub for this. That's how Integral-Calculator.com solves it. I think your teacher just doesn't want you to solve it this way to practice polynomial division. – Zaya Mar 05 '20 at 00:26
  • @GEdgar Oh I did include that, I just factored out the constant from the integral. – mohhamed rafi Mar 05 '20 at 00:28

1 Answers1

3

Dividing top & bottom by $2$ and fairly directly integrating your original integral gives

$$\begin{equation}\begin{aligned} \int \frac{2x}{2x+4} dx & = \int \frac{x}{x+2} dx \\ & = \int \frac{x + 2 - 2}{x+2} dx \\ & = \int \left(1 - \frac{2}{x + 2}\right) dx \\ & = x - 2\ln|x + 2| + C_1 \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Note I effectively did polynomial division where I used that $x = x + 2 - 2$ in the numerator in the second line. Next, using your substitution of $u = 2x + 4$ gives $du = 2dx \implies dx = \frac{du}{2}$. Thus, you then get

$$\begin{equation}\begin{aligned} \int \frac{2x}{2x+4} dx & = \frac{1}{2}\int \frac{u - 4}{u}du \\ & = \frac{1}{2}\int \left(1 - \frac{4}{u}\right)du \\ & = \frac{1}{2}\left(u - 4\ln|u|\right) + C_2 \\ & = \frac{1}{2}\left((2x + 4) - 4\ln|2x + 4|\right) + C_2 \\ & = x + 2 - 2\ln|2(x + 2)| + C_2 \\ & = x - 2(\ln|x + 2| + \ln(2)) + 2 + C_2 \\ & = x - 2\ln|x + 2| + (2\ln(2) + 2 + C_2) \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

Note you made a small mistake in your question text where you used $dx$ instead of $du$. As you can see, you get the same result, with $C_1$ from \eqref{eq1A} being equal to $2\ln(2) + 2 + C_2 = \ln(4) + 2 + C_2$ from \eqref{eq2A}, so using substitution does work.

There's no reason why you can't use substitution. Instead, as several question comments have indicated, it's fairly likely your review notes stated to use polynomial division for you to practice that technique instead.

John Omielan
  • 47,976