Integration by substitution and by partial fractions lead to different results When integrating $\int \frac{3x-2}{x+1}dx$ we can take two paths.
$\alpha)$ Let $u=x+1$, so
$$\int \frac{3x-2}{x+1}dx = \int \frac{3u-5}{u}du = 3u- 5\ln|u|=3(x+1)-5\ln|x+1|$$
$\beta)$ See that $3x-2=3(x+1)-5 \implies \frac{3x-2}{x+1}=3-\frac{5}{x+1}$. Then $$\int \frac{3x-2}{x+1}dx=\int (3-\frac{5}{x+1})dx = 3x-5\ln|x+1|$$
Clearly, the two results are very similar. They only differ in that $x \neq x+1$. Of course I must be making a mistake in some of the approaches (or both), but I can not seem to find where the error is. Anybody care to point it out for me? Thanks in advance.

Both of them are correct, because you forget the $+C$ part.
$$\begin{align}\int \frac{3x-2}{x+1}dx&=3(x+1)-5\ln|x+1|+C_1\\
\\
\int \frac{3x-2}{x+1}dx&=3x-5\ln|x+1|+C_2\end{align}$$
Since $C_1$ and $C_2$ are arbitrary constants, if you define $C_2=3+C_1$, they are the same.
