# Integral Help

1. Aug 18, 2009

### Stratosphere

1. The problem statement, all variables and given/known data
As I was reviewing some of my previuosly learned calculus I came across somthing that I had either forgoten how to do or was never taught. How do you take the integral of somthing like this

$$\int \sqrt{\frac{9}{4}x+1}$$

I don't know how to start with this one. Do I use U substitution?

2. Aug 18, 2009

### CFDFEAGURU

First rewrite the integral as

dy=$$\int(\frac{9}{4x}+1)^{1/2}$$ dx

Now use the u substitution.

Thanks
Matt

3. Aug 18, 2009

### Stratosphere

Using U substitution I got ((9/4x+1)^1/2)/(27/8) I don't think I did this right.

4. Aug 18, 2009

### CFDFEAGURU

I don't believe that is correct.

Try this.

Let u = $$\frac{9}{4x}$$+1

Now find du/dx.

5. Aug 18, 2009

### Stratosphere

Du/dx is =36/16x$$^{2}$$

6. Aug 18, 2009

### CFDFEAGURU

Can you show your steps as to how you got that result please?

7. Aug 18, 2009

### Stratosphere

u=$$\frac{9}{4x}+1$$

$$\frac{0*4x-9*4}{(4x)^{2}}$$ So I made a mistake with the negative? The answer should be -36/16x^2.

8. Aug 18, 2009

### CFDFEAGURU

Yes, you got it. Now perform the rest of the u substitution procedure and your done.

9. Aug 18, 2009

### Elucidus

I think the discussion got off track somewhere. The radicand above is $\frac{9}{4}x+1$ not $\frac{9}{4x}+1$. Or was the original statement an error?

If the original is correct, then it can be easilly integrated by substitution - try $u=\frac{9}{4}x+1$.

(This looks like a Stewart's Calculus exercise.)

--Elucidus