This function has two different integrals?

Question

$f(x)=∫\frac{1}{x^2}dx$

Integrating by u-substitution:

$u=x^2$

$du=2dx$

$\frac{1}{2}du = dx$

$∫\frac{1}{x^2}dx=$ $∫\frac{1}{u}\times\frac{1}{2}du$

$\frac{1}{2}$∫ $\frac{1}{u}du$

$=\frac{1}{2}ln u+c$

$=\frac{1}{2}ln x^2+c$

$=lnx+c$

Another way:

$∫\frac{1}{x^2}dx=∫x^{-2}dx $

$∫x^{-2}dx$

$=\frac{x^{-1}}{-1} + c$

$=-\frac{1}{x} + c$

Where have I gone wrong?

Answer 1

Your mistake is in changing the $dx$: it should be $du = 2x \, dx$, or $dx = \frac{1}{2}u^{-1/2} \, du$, which gives $$ \int \frac{dx}{x^2} = \int \frac{1}{u} \frac{1}{2u^{1/2}} \, du = \frac{1}{2} \int u^{-3/2} \, du = -u^{-1/2}+C = -\frac{1}{x}+C. $$

Answer 2

It’s not $du=2dx$, it’s $du=2xdx$.

Answer 3

If $u=x^2$, then $\mathrm du=2x\,\mathrm dx$.