# Indefinite Integrals?

Need help finding the anti derivatives of $\dfrac{\sqrt{x}}{3}$ and $\dfrac{-3}{5x}$.

I don't know what to do when the problem is in a fraction.

• If $F(x)$ is an antiderivative of $f(x)$, then $aF(x)$ is an antiderivative of $af(x)$. – egreg Dec 3 '13 at 23:54

You an take out the constant (Constant Factor Rule) and integrate normally:

$$\int \frac{\sqrt x}{3} = \frac{1}{3} \int \sqrt x = \frac{1}{3} \cdot \frac{2x^{3/2}}{3} = \frac{2x^{3/2}}{9}$$

$$\int -\frac{3}{5x} = -\frac{3}{5} \int \frac{1}{x} = -\frac{3}{5} \ln |x|$$

We have constants, and we have the variable $x$.

$$\frac {\sqrt x}{3} = \dfrac 13 \sqrt x = \dfrac 13 x^{1/2}$$

Use the power-rule, and the fact that $\int (af(x))\,dx = a \int f(x)\,dx$.

Now, for the next, we have $$\dfrac{-3}{5x} = \dfrac{-3}5\cdot \dfrac 1x$$ and recall that $$\dfrac {d}{dx}\left(\ln x\right) = \dfrac 1x$$