Find $ \int {dt\over 2t+1}$ a simple question, but I'm stuck anyway:
How to integrate this:
$$ \int {dt\over 2t+1}  = ? $$
Is it simply: $\ln|2t +1| $
or do I need Chain rule like: $$\ln|2t +1| \cdot \frac{d}{dt}(2t + 1) $$
 A: You've got the right idea about the "form" of the integral, but recall, we need to account for the chain-rule before integrating, by u-substitution for example.
In general, when you have an integral of the form $\int \dfrac{f'(t)}{f(t)} \,dt$
your result will indeed be: $$\int \dfrac{f'(t)}{f(t)} \,dx = \ln|f(t)| + C$$
In your case, we have $f(t) = u =2t + 1$. Now, we need $f'(t):\; du = 2 dt\iff \frac 12 du = dt$, so we need to obtain the form $$\int \frac{f'(t)}{f(t)}\,dt$$ which we can obtain directly, or through substitution:
$$\int \frac{dt}{2t + 1} = \frac 12 \int \underbrace{\frac{2\,dt}{2t + 1}}_{\dfrac{f't}{f(t)} dt} \quad \overset{\text{substitute}}{=} \quad \frac 12 \int \frac{du}{u} \quad = \quad  \frac 12\ln|u| + C = \cdots$$
A: This should be the straight-cut way to integrate the expression.
As Cameron Buie responded, let $u = 2t + 1$.  Then, $du = 2 dt$.  Making that substitution for the integral, we have:
$$\int \frac{du}{2u}$$
$$= \frac{1}{2} \int \frac{du}{u}$$
$$= \frac{1}{2} \ln|u| + \mathrm{c}$$
$$= \frac{1}{2} \ln|2t + 1| + \mathrm{c}$$
A: Hint: Use the substitution $u=2t+1$, and see what happens.
A: Hint: Let $u=2t+1$. I think it will help you.
