$\int \frac{x}{x+1}\,dx$ can't find mistake in a step

Question

$\int \frac{x}{x+1}\,dx$ so I set substitution to $x+1=t$ then I differentiate to get $dx=dt$ and then proceed to get $\int 1\,dt - \int \frac{1\,dt}{t}$ to get $t-\ln(|t|)+C$ but when I replace back the substitution I get $x+1$ in the final answer instead of just the $x$ , so my question is why was it a mistake to integrate $\int 1\,dt$ instead to switch back to x and then integrate $\int 1\,dx$

Answer 1

There is nothing wrong in your solution .

$x+1-\ln|x+1| + C$

Now let $C+1=C'$

$x-\ln|x+1|+C'$

You can always adjust the constant of integration.

Answer 2

Alternative:

$\int \frac{x}{x+1}\, dx=\int \frac{x+1-1}{x+1}\, dx=\int 1-\frac{1}{x+1}\, dx= x-\ln|x+1|+C$

Without substitution.