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

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