Integrating Difficult Equations: U Substitution

[Stratosphere]

Homework Statement

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?

 

[CFDFEAGURU]

First rewrite the integral as

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

Now use the u substitution.

Thanks
Matt

 

[Stratosphere]

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

 

[CFDFEAGURU]

I don't believe that is correct.

Try this.

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

Now find du/dx.

 

[Stratosphere]

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

 

[CFDFEAGURU]

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

 

[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.

 

[CFDFEAGURU]

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

 

[Elucidus]

Homework Statement

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?

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