0
$\begingroup$

The given problem is: \begin{aligned} \int {x^{1/2}\over {x^{1/2}} - 3} dx \end{aligned}

The textbook (Larson, Edwards) 9th edition suggests to use u-substitution and let u be the denominator.

So, I made $u =x^{1/2} -3$. So, $u+3 = x^{1/2}$ Thus, du is $\frac{1}{2\sqrt{x}}$. So, $2x^{1/2} du = dx$, and that follow $2(u+3) du = dx$.

So, I then put the original integral in terms of u: \begin{aligned} 2\int {u+3{}\over {u}} du \end{aligned}

When I carry out the integration, I do not get the correct answer. I noticed something was a bit fishy when I moved the u up and expanded to get $u^0 + 3u^{-1}$ which seemed a bit odd for this problem. I expect something to be wrong with my numerator when rewriting the function, but I cannot seem to find any errors.

Any help would be appreciated.

  • 3
    Thanks for cleaning up my post Emile with LaTeX.2012-03-01

3 Answers 3

2

You have to substitute for the $\sqrt x-3$ in the denominator, the $\sqrt x$ in the numerator, and the $dx$. I think you only did two of the three.

  • 0
    @AndréNicolas Well, I noticed the obvious$u+3$for x^1/2, but I was just having troubles with the (u+3) term also for going from dx to du as well. Nonetheless, thanks.2012-03-01
0

As far as I can tell you are on the right path, once you took u back out of the result what did you get? I don't think you've made any mistakes...

  • 0
    You might want to reconsider.2012-03-02
-1

Your problem appears to be in changing over from x to u.

$\int \dfrac{x^\frac12}{x^\frac12-3}dx=2\int\dfrac{x}{x^\frac12-3}\left(\dfrac{x^{-\frac12}dx}2 \right)=$

$\int\dfrac{(u+3)^2}{u}du$

  • 0
    Not sure what you mean too elliptical, but I used the results that the OP already calculated. He determined du correctly and he had a formula for $$x$^\$f$rac12$. All I did was clearl$y$ separate out the du and per$f$orm the proper substitutions. It should be easily solvable from there.2012-03-02