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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.

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    $\begingroup$ Thanks for cleaning up my post Emile with LaTeX. $\endgroup$
    – Joe
    Commented Mar 1, 2012 at 22:24

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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.

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    $\begingroup$ It appears I have missed the part about substituting in for dx. Can you elaborate? I think I understand what you mean - since I only threw a 2 out in front of the integral, I also missed the (u+3) term needed since 2(u+3) du = dx? Thus, it would be 2 times the integral of (u+3)^2/(u) du? $\endgroup$
    – Joe
    Commented Mar 1, 2012 at 22:22
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    $\begingroup$ @Jay The integral in this comment is correct. The integral in the answer is not. Also, notice that when you expand $(u+3)^2$ it is easily solvable. $\endgroup$
    – Artium
    Commented Mar 1, 2012 at 22:29
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    $\begingroup$ It is not absolutely clear which one you missed. But most likely from what you wrote is that you missed substituting $u+3$ for the $x^{1/2}$ that was originally on top. Anyway, the expression in your comment above is right. $\endgroup$
    – André Nicolas
    Commented Mar 1, 2012 at 22:31
  • $\begingroup$ @Artium Yup, thanks. $\endgroup$
    – Joe
    Commented Mar 1, 2012 at 22:35
  • $\begingroup$ @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. $\endgroup$
    – Joe
    Commented Mar 1, 2012 at 22:36
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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...

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  • $\begingroup$ You might want to reconsider. $\endgroup$
    – Gerry Myerson
    Commented Mar 2, 2012 at 0:16
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Your problem appears to be in changing over from x to u.

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

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

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  • $\begingroup$ Someone care to elaborate the reason for the downvote? $\endgroup$
    – Mike
    Commented Mar 1, 2012 at 22:45
  • $\begingroup$ I didn't downvote, but, given the level of the question, I suspect this answer is too elliptical to be very helpful. $\endgroup$
    – Gerry Myerson
    Commented Mar 2, 2012 at 0:16
  • $\begingroup$ 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 $x^\frac12$. All I did was clearly separate out the du and perform the proper substitutions. It should be easily solvable from there. $\endgroup$
    – Mike
    Commented Mar 2, 2012 at 0:25

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