# Verification neededproblem seems odd

1. Oct 7, 2007

### gochi

Hi,

I have to find the integral of 1/square root(x-3)dx

this is what i did

let u=sqr root(x-3)
du=1/2*square root(x-3)dx
dx=2*square root(x-3)du

the integral of 1/square root(x-3)dx= 2*square root(x-3)(1/u)du

..ok now what i did is plug back u= square root of(x-3) in the denom
that way Im only left with 2du.

Then i integrated that to 2x2/2 which is equal to x^2.

Ultimatley I got the integral to equal X !!!! which cant be right!!!

But doesnt this make sense, I know i did not do anything wrong...

2. Oct 7, 2007

### neutrino

$$\frac{d}{dx}\left(\sqrt{x-3}\right) \neq \frac{1}{2}\left(\sqrt{x-3}\right)$$

3. Oct 7, 2007

### SiddharthM

..ok now what i did is plug back u= square root of(x-3) in the denom
that way Im only left with 2du.

that move is NOT legit.

the integral of 1/square root(x-3)dx= 2*square root(x-3)(1/u)du

neither is that one, wen u change variables, ur new equation must have only that variable - or constants of course...if you write root(x-3) IN TERMS of u that is fine, but the way it is written doesn't make sense.

Hint: let u=x-3.

4. Oct 7, 2007

### gochi

I meant that the derivitive of sqroot(x-3)=1/2sqroot(x-3)

5. Oct 7, 2007

### neutrino

And I'm saying that it is not.

$$\frac{d}{dx}\left({x-3}\right)^\frac{1}{2} = \frac{1}{2}\left(\frac{1}{\sqrt{x-3}}\right)$$

6. Oct 7, 2007

### SiddharthM

"1/2sqroot(x-3)", i think this is 1 OVER 2 times the square root of x minus 3.

7. Oct 7, 2007

### atqamar

Your first part is flawless. Integrate by parts:
$$\int xdu = xu-\int udx$$
set $$x=2$$ and $$u=\sqrt{x-3}$$. Therefore $$du=\frac{1}{2\sqrt{x-3}}$$. Obviously $$dx=0$$.
$$\int 2(\frac{1}{2\sqrt{x-3}}) = 2(\sqrt{x-3})-\int 0$$

Simplify and you get:
$$\int \frac{1}{\sqrt{x-3}} = 2\sqrt{x-3}$$

As you do more integrations, this process will become easier.

8. Oct 7, 2007

### SiddharthM

u-substituting x-3 is a lot easier!

9. Oct 7, 2007

### HallsofIvy

Staff Emeritus
The problem is that most people would interpret it as (1/2) sqroot(x-3), not 1/(2 sqroot(x-3)). USE PARENTHESES!

10. Oct 7, 2007

### SiddharthM

that he didn't make that mistake is clear. Look at his next step:

dx=2*square root(x-3)du

11. Oct 8, 2007

### neutrino

Ah... I apologise, gochi. But as Halls emphasised, please make use of parentheses. :)