$$\int \frac {dx} {x(x^8-256)}$$
I am supposed to use the formula
$$\int \frac {dx} {x(ax+b)} = \frac1b\ln\left|\frac x {ax+b}\right|+C $$
to find the integral.
I don't know how to start. Help is appreciated.
$$\int \frac {dx} {x(x^8-256)}$$
I am supposed to use the formula
$$\int \frac {dx} {x(ax+b)} = \frac1b\ln\left|\frac x {ax+b}\right|+C $$
to find the integral.
I don't know how to start. Help is appreciated.
On
"How to start" is, obviously the $x$ in your problem, $$\int \frac {dx} {x(x^8 - 256)},$$ cannot be the same $x$ as the $x$ in $$\int \frac {dx} {x(ax+b)}.$$
So you will have to make some kind of variable substitution in your problem integral, and hope to end up with something that looks like $$\int \frac {dy} {y(ay+b)}.$$
So we look for patterns in this that match your problem, and we see that $(ay + b)$ looks like $(x^8 - 256)$ if we set $ay = x^8$ and $b = 256$. Better still, let's set $a = 1$ and $y = x^8$.
From that point on, follow the solution given by @prrulz.
Note that it is really not obvious (to some of us, anyway) that the substitution $y = x^8$ will work, but it seems like a plausible thing to try. So you try it, and find that it works. Unless you have some special wiring in your brain that helps you instantly see the final result of any substitution, you'll just have to try and see sometimes.
Make the substitution of $$y = x^8.$$
Then $$dy = 8x^7dx \Rightarrow \frac{dy}{8x^7} = dx.$$
So $$\int\dfrac{dx}{x(x^8 - 256)} = \int\dfrac{dy}{8x^7}\dfrac{1}{x(y - 256)} = \int\dfrac{dy}{8y(y - 256)} = \frac{1}{8}\left(\frac{1}{-256}ln|\frac{y}{y - 256}|\right) + C = \frac{1}{8}\left(\frac{1}{-256}ln\left|\frac{x^8}{x^8 - 256}\right|\right) + C$$