Integrating by substitution containing a further factor problem

Question

The problem:

$$ {\int } (x+1)(3x+1)^9 dx $$

let u = 3 x +1

3 x = u - 1

$ x = \frac{1}{3} (u-1) $

Hence,

$ x + 1 = \frac{1}{3} (u-1) + 1$

$ = \frac{1}{3} (u+2) $ This line here I do not understand.

The + 1 become a 3? How does this happen?

Answer 1

Hint:

$$ x+1=\frac{1}{3}(u-1)+1 \quad \iff \quad x+1=\frac{1}{3}u -\frac{1}{3}+1=\frac{1}{3}u+\frac{2}{3} $$