Evaluate $\int \frac{1}{ax^2-bx}\,dx$ with substitution

Question

Evaluate $\int\frac{1}{ax^2-bx}\,dx$

First try:

$$\int\frac{1}{ax^2-bx}\,dx = \int\frac{1}{bx^2(\frac{a}{b}-\frac{1}{x})}\,dx$$

By substituting $u=\frac{a}{b}-\frac{1}{x}$ so $du=\frac{1}{x^2}\,dx$ we have,

$$\int\frac{1}{bx^2(\frac{a}{b}-\frac{1}{x})} \, dx = \frac{1}{b} \int\frac{1}{u} \, du = \frac{\ln|u|}{b}+C = \frac{\ln|\frac{a}{b}-\frac{1}{x}|}{b}+C=\frac{\ln|\frac{ax-b}{bx}|}{b}+C$$

Second try:

$$\int\frac{1}{ax^2-bx} \, dx = \frac{1}{b}\int\frac{b}{x^2(a-\frac{b}{x})} \, dx$$

By substituting $u=a-\frac{b}{x}$ so $du=\frac{b}{x^2}dx$ we have,

$$\frac{1}{b}\int\frac{b}{x^2(a-\frac{b}{x})}dx = \frac{1}{b}\int\frac{1}{u}\,du = \frac{\ln|u|}{b}+C = \frac{\ln|a-\frac{b}{x}|}{b}+C=\frac{\ln|\frac{ax-b}{x}|}{b} + C$$

It seems to me that $\frac{\ln|\frac{ax-b}{bx}|}{b}+C\neq\frac{\ln|\frac{ax-b}{x}|}{b}+C$, so, whats the matter ?!

Answer 1

Note that

$$\frac{\ln|\frac{ax-b}{bx}|}{b}+C=\frac{\ln|\frac{ax-b}{x}|}{b} + \frac{\ln |\frac{1}{b}|}{b} + C=\frac{\ln|\frac{ax-b}{x}|}{b}+C_2$$

Answer 2

$$\frac{1}{b}\ln\left|\frac{ax-b}{bx}\right|=\frac{1}{b}\left(\ln\left|\frac{ax-b}{x}\right|-\ln|b|\right)=\frac{1}{b}\ln\left|\frac{ax-b}{x}\right|+\frac{1}{b}\ln|b|$$ And $\frac{\ln|b|}{b}$ is just a constant.

The primitive function is not just a function - it's a set of functions: $$\int f :=\{g\mid g'=f\}$$ And we usually denote this set as one of its element + a constant. So, for example, we can say that $\int x = \frac{x^2}{2}+C$, or $\int x = \frac{x^2}{2}+2-e^\pi+C$, they denote the same set.

Note: Our calculus teacher was not using the $+C$. You can also leave it out, but it might cause some problem later, for example, when you will deal with differential equations.