Anti derivative of sqrt{2x+1}

1. May 24, 2005

ktpr2

I know the anti derivative of $$\sqrt{2x+1}$$ is $$1/3(2x+1)^(3/2)$$ but I can't develop a concrete method for finding the anti derivative of functions like this. How would you go about finding the anti derivative of this function?

2. May 24, 2005

dextercioby

Make the substitution

$$2x+1=u$$

Daniel.

3. May 24, 2005

Jameson

Right. To go further with that substitution method...

$$\int \sqrt{2x+1}dx$$

$$u = 2x+1$$

$$du = 2dx$$

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

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

Take it from there.

Jameson

Last edited: May 25, 2005
4. May 24, 2005

ktpr2

with $$2x+1=u$$
you get $$\sqrt{u} = u^(^1^/^2^)$$
and using the $$x^n -> \frac{x^n^+^1}{n+1}$$ rule I get $$2/3 (u)^(^3^/^2^)$$ which is $$2/3 (2x+1)^(^3^/^2^)$$ which isn't correct, so what am i missing here?

[edit- oh. I was trying to work with the take of anti derivatives to come up with the answer so i could use it to integrate, but it looks like you have to use a little integration sometimes to get antiderivatives (indefinte integrals) sometimes. thanks.]

Last edited: May 24, 2005
5. May 24, 2005

Jameson

Look at my post... you need to factor out that one-half.

6. May 24, 2005

whozum

An even better/more appropriate excercise: Expand it to the family

$$\int \sqrt{ax+b} \ dx$$

$$u = ax+b, du = a dx$$

$$\frac{1}{a} \int \sqrt{u} \ du$$

$$\frac{1}{a} \ \frac{2}{3}\ u^{\frac{3}{2}}$$

$$\frac{2}{3a} \ u^{\frac{3}{2}}$$

With a = 2, b = 3 you get

$$\frac{1}{3} \ u^{\frac{3}{2}}$$

Last edited: May 24, 2005
7. May 24, 2005

Integral

Staff Emeritus
Jameson,
$$\int \sqrt{2x+1}dx = \frac{1}{2}\int udu$$

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

8. May 25, 2005

Jameson

Ah yes, an important typo. I'll fix it. Thanks.

9. May 25, 2005

Jameson

The anti-derivative and the integral are the same thing... I'm confused about what you mean.