Integrate $\frac{1}{\sqrt{x}}$

Question

I have recently started an integration unit, and im just now getting familiar eith the concept of integrals. However, I cant figure out why $$\int\frac{1}{\sqrt{x}}dx=2\sqrt{x}+C$$ Wolfram Alpha doesnt show any steps for it, and I cant seem to work it out on paper. Any help would be appreciated!

Answer 1

You can use that $$\int x^ndx=\frac{x^{n+1}}{n+1}+C$$ if $$n\ne -1$$ and write your integral as

$$\int x^{-1/2}dx$$