Regarding $\int \frac{4x}{4x-3}\mathrm dx$

Question

Integrate $\frac{4x}{4x-3}$

What I did was split them into two separate fractions. So, I had

$\int\left(1 + \frac{3}{4x-3}\right) dx$. Then, I had $x + \frac{3}{4}\ln(4x-3) + C$ as an answer. But, in the mark scheme, it said something like this:

By the way, in this example, I think $\log$ means "$\ln$". Can you please show working out that gives you a final answer same as above?

Answer 1

Let $u = 4x - 3 \implies du = 4\,dx \iff dx = \frac 14 du$

Note that $4x = u+3$.

That gives you $$\frac 14\int \frac{u+3}{u} \,du = \frac 14\left(\int \,du) + 3\int \frac{du}{u}\right) = \frac 14u + \frac 34\ln|u| + C $$

Now replace $u = 4x-3$ to get $$\frac 14(4x-3) + \frac 34 \ln|4x-3| + C = x + \frac 34\ln(4x-3) +\underbrace{ (C -\frac 34)}_{C'}$$

You see that your answer is equivalent to the text's answer, up to a constant.

Answer 2

Ah, but you only disagree on an additive constant. Your solution is general, but the mark scheme has chosen a specific value for $C$. Perhaps your function has to go through some particular point?