Indefinite integral fractional roots

Question

I know the result of these indefinite integrals, but I don´t understand how the calculaton gets there:

$$\int \frac{1}{\sqrt{x}}dx = 2 \sqrt{x}$$

$$\int \frac{1}{\sqrt[3]{x}}dx = \frac{3x^{\frac{2}{3}}}{2}$$

In the first case, considering x=2t I get into :

$$\int \frac{1}{\sqrt{x}}dx = \int \frac{1}{t}2tdt = 2\int 1dt = ?? $$

But I don´t know how to follow from there.

Answer 1

Hints :

• Use $\frac{1}{\sqrt{x}}=x^{\frac{-1}{2}}$
• The second integrand is equal to $x^{\frac{-1}{3}}$
• The antiderivate of $x^n$ is $\frac{x^{n+1}}{n+1}$ for every real $n\ne -1$