Evaluating the indefinite integral I am having trouble understanding this homework question:
$$\int \frac {dx}{cx+h} $$
So, what I thought I should do is...
$$\int \frac {dx}{cx+h} $$
let $u$ be: $cx+h$
let $du$ be: $ 1\,dx $
$$\int \frac {du}{u} = \int u^{-1}\,du $$
$$\int u^{-1}\,du = \ln(u) + C = \ln(cx+h) + C $$
$$ \ln(cx+h) + C $$
But, I got it wrong, what am I doing wrong?
 A: I really don't think a change of variables is necessary here, in my opinion it kind of clouds things up. Integrating is finding an antiderivative, and when we can't, we manipulate the expression to see if it's recognizably the derivative of something. When we can't do this we ask Wolfram Alpha (jest). So, since we know $$(\log x)'=\frac1x$$
We might say that what we have is the derivative of $$\log(cx+h)
$$
But if we take the derivative we obtain $$\frac{c}{cx+h}$$
This is very close to our integral! We can modify it, though (without actually modifying it) to make it look as such: $$\int{\frac{1}{cx+h}}=\int{\frac1c\frac{c}{cx+h}}=\frac1c\int{\frac{c}{cx+h}}=\frac1c\log(cx+h)+k$$
I'd say as a rule of thumb always look for "this function looks like the derivative of something or at least almost", before plugging in any change of variables!
A: You are attempting integration by substitution, and make the correct substitution $u = cx + h$.
Now where you go wrong is when you say "let $du = dx$". The key here is that although two differentials, such as $du$ and $dx$ are both very "small", they are not necessarily equal; if they where equal then $\frac{du}{dx}$ would equal $1$ which would mean that $u$ changes by the same amount when you change $x$, but I'm sure you can differentiate $u$ and will find actually that $\frac{du}{dx} = c$. Now $\frac{du}{dx}$ is just a fraction, so it must be that $du$ is actually bigger than $dx$, $c$ times bigger in fact, which is is why you must use the substitution $dx = \frac{1}{c}dx$ when you substitute $u$ for $x$ in the original equation.
A word of warning: although you can think of $\frac{du}{dx}$ as a normal fraction, it only "works" because both $du$ and $dx$ are differentials and very small (although not equal!), you can't divide any old variable by a differential and get a valid result.
A: There is just one mistake,
$$ \mathrm{d}u=c \mathrm{d}x $$
Then the answer to,
$$ \displaystyle\ \int {\dfrac{1}{cx+h}} \mathrm{d}x $$
is,
$$ \dfrac{1}{c} \ln| cx+h | $$
