When to take log in integrals?

Question

When is log taken in integrals and when its not and why? which of the below is correct and why?

$$\int {1\over\sqrt{t}} dt = \log |\sqrt{t}| + C$$

or

$$\int {1\over\sqrt{t}}dt = 2\sqrt{t} + C$$

Answer 1

Firstly, $$\int \frac{dt}{t}=\ln|t|+c$$

You don't 'take logs' when you're integrating per se, rather you use techniques like substitution, or algebraic manipulation, trig rules etc, to change an integral you don't know into an integral you do know.

Here's an example of algebraic manipulation:

$$\int \frac{dt}{\sqrt{t}}=\int \frac{dt}{t^{1/2}}=\int t^{-1/2}\,dt=\frac{t^{1/2}}{1/2}+c=2t^{1/2}+c$$

Here we used the Fundamental Theorem of Calculus to know $$\int t^n \,dt=\frac{t^{n+1}}{n+1}+c\quad \text{for } n\neq -1 $$

On the other hand suppose you want to find

$$\int \tan(t) \,dt=\int \frac{\sin(t)}{\cos(t)}\,dt$$

If you don't know offhand, then you might try a substitution $u=\cos(t)$, $du=-\sin(t) dt$. Thus

$$\int \frac{\sin(t)}{\cos(t)}\,dt=-\int \frac{du}{u} $$

Now it's in the form we know for the logarithm so

$$\int \tan(t) \,dt=\int \frac{\sin(t)}{\cos(t)}\,dt =-\int \frac{du}{u}=-\ln|u|+c=-\ln|\cos(t)|+c$$

We can check if we're right,

$$\frac{d}{dt}\ln|f(t)|=\frac{f'(t)}{f(t)}$$

So,

$$\frac{d}{dt}\left(-\ln|\cos(t)|\right)=-\frac{-\sin(t)}{\cos(t)}=\tan(t)$$

To summarise, in general

$$\int \frac{dt}{f(t)}\color{red}{\neq} \ln|f(t)|+c$$

For some examples to back this up there's a nice table at Pauls online notes