1. Nov 4, 2005

How do I do this?
$$\int {\frac {1} {(\sqrt {-x})}} dx$$
I got $$2 \sqrt {-x}$$ but my teacher got $$-2 \sqrt {-x}$$ and I don't know how he got there.

Last edited: Nov 4, 2005
2. Nov 4, 2005

### Tom Mattson

Staff Emeritus

3. Nov 4, 2005

$$\int {\frac {1} {\sqrt{-x}}} dx = \int {\frac {1} {(-x)^{1/2}} dx = \int {(-x)^{-1/2}} dx = \frac {(-x)^{-1/2+1}} {-1/2+1} = \frac {(-x)^{1/2}} {1/2} = 2 \sqrt {-x}$$

Last edited: Nov 4, 2005
4. Nov 4, 2005

### VietDao29

Nope,
You are wrong when going from:
$$\int (-x) ^ {-\frac{1}{2}} dx$$ to $$\frac{(-x) ^ {-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} + C$$.
The reason is that, you only have:
$$\int x ^ \alpha dx = \frac{x ^ {\alpha + 1}}{\alpha + 1} + C$$.
You do not have:
$$\int (-x) ^ \alpha dx = \frac{(-x) ^ {\alpha + 1}}{\alpha + 1} + C$$
So, from $$\int (-x) ^ {-\frac{1}{2}} dx$$, you can use u-substitution:
Let u = -x, so du = -dx (or you can say dx = -du). So the integral becomes:
$$\int u ^ {-\frac{1}{2}}(-du) = - \int u ^ {-\frac{1}{2}}du = -2 u ^ {\frac{1}{2}} + C$$. Since u = -x, so change u back to x, you'll have:
$$-2\sqrt{-x} + C$$