Compute $\int\frac{x}{x+1}dx$

Question

My textbook uses the following method:

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

It seemed obvious after seeing it solved, but I didn't spot that first step of splitting the fraction.

So I attempted it using u substitution:

$u = x+1$
$x = u-1$
$dx = du$

$$\int \frac{u-1}{u} du = \int 1-\frac{1}{u}=u-\ln|u| = (x+1) - \ln|x+1|$$

The two expressions after using subsitution are equivalent aren't they?

$$\int \frac{u-1}{u} = \int \frac{x +1-1}{x+1}$$

So when doing the u substitution, it appears that I'm getting to the same stage where they split the fraction. Can anyone explain what I'm doing wrong?

Thanks in advance

Answer 1

Note that

$$\int \frac{x}{x+1} dx = \int \frac{x +1-1}{x+1}dx=\int \left(1-\frac{1}{x+1}\right)dx=x-\ln|x+1|\color{red}{+C}$$

Answer 2

The two expressions you listed above ie (x+1) - log(x+1) +C and x - log(x+1)+C, are in fact the same and equal because, in the first expression you can simply consider the (+1) as a part of the arbitrary constant C.

Since, the indefinite integrals represent the family of curves and not the exact equation, BOTH EXPRESSIONS SHALL BE CONSIDERED OF THE SAME FAMILY and hence the same. Thanks for asking mate!