Integration by partial fraction decomposition

Question

I have the following example question in my homework and for the life of me I cannot understand the second to last step which I circled in red. Why are they suddenly changing the factors of 1/4 to 1/12 and -1/12, and the numerators to 3 and -3?

Answer 1

So, we know $3/12 = 1/4$, that much is easy to understand.

What happens between the last two lines is kinda skipped over since once you get used to it, we kinda skip the details.

So first we have those $1/4$ coefficients. We move them outside the integral. Next, we divide them by $3$ and multiply the inside of the integral by $3$. Why do we do that? It's because the integrals that result lend themselves to a nice $u$-substitution.

In the left integral, we have $u = 2+3x,$ and $du = 3dx$.

In the right integral, we have $u = 2-3x, du = -3dx$.

Each integral then becomes of the form $\int du/u$, which has antiderivative $\ln|u|$. And then we just bring in the $u$ again.

Generally, if the numerator of a fraction in an integral like this is the derivative of the denominator, then the antiderivative is simply the natural logarithm of the denominator. That's essentially what happened here - we just made a manipulation to make that the case.

Answer 2

The idea is that you want to have a 3 in the numerator as it makes the integral a bit easier to calculate.

Observe that $\int \frac{f'(x)}{f(x)}dx=\log(f(x))+C$.

Therefore, $\int \frac{3}{2+3x}dx=\log|2+3x|+C$.

Similarly for the other term.

Answer 3

The chain rule is the reason for factoring $3$ out. Note $\frac{\operatorname{d}}{\operatorname {dx}}\ln\mid2+3x\mid=\frac{\color{blue}{3}}{2+3x}$.