If $u^2 = 2x-1$, why does $du^2 = dx$?

Question

So the problem i encountered was:

$$\int_1^5\frac{x}{\sqrt {2x-1}}dx$$

the first step in the given solution was:

$$u^2 = 2x-1 \Rightarrow 2du^2 = 2dx \Rightarrow udu ~\text{(or} ~ du^2\text{)} = dx,$$

which confused me, as the derivative of $u^2$ in this situation should be $2 dx$.

Tiny recommendation: it's a really good idea to write $d(u^2)$ in circumstances like this so it's not confused with $(du)^2$ (which is a depressingly easy mistake to make in this sort of problem). — Steven Stadnicki, Mar 06 '17 at 20:44

score 3 · Accepted Answer · answered Mar 06 '17 at 20:42

$$\textrm{d}u^2 = 2u\ \textrm{d}u \\ \textrm{d}(2x-1) = 2\ \textrm{d}x$$ Because $u^2=2x-1$, we know that $\textrm{d}u^2 = \textrm{d}(2x-1)$ as well. And thus $2u\ \textrm{d}u = 2\ \textrm{d}x$, or cancelling the $2$'s: $$u\ \textrm{d}u = \textrm{d}x$$

If your solution sheet says anything else, it must be a typo as Wojowu says in the comments.

score 1 · Answer 2 · answered Mar 06 '17 at 20:46

1

$\int_1^5\frac{x}{\sqrt {2x-1}}dx$

$u^2 = 2x -1\\ x = \frac 12(u^2 + 1)\\ dx = u\ du\\$

$\int_1^3 \frac 12 (u^2+1) du$

answered Mar 06 '17 at 20:46

Doug M

57,877

If $u^2 = 2x-1$, why does $du^2 = dx$?

2 Answers2