How Do You Find the Antiderivative of sqrt(2x+1)?

[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?

 

[dextercioby]

Make the substitution

2x+1=u

Daniel.

 

[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

 

[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.]

 

[Jameson]

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

 

[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}}

 

[Integral]

Jameson,
Your line:
\int \sqrt{2x+1}dx = \frac{1}{2}\int udu

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

 

[Jameson]

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

 

[Jameson]

[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.]

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