Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
3  
Thanks for cleaning up my post Emile with LaTeX. –  Joe Mar 1 '12 at 22:24
add comment

3 Answers

up vote 2 down vote accepted

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.

share|improve this answer
1  
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? –  Joe Mar 1 '12 at 22:22
1  
@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. –  Artium Mar 1 '12 at 22:29
2  
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. –  André Nicolas Mar 1 '12 at 22:31
    
@Artium Yup, thanks. –  Joe Mar 1 '12 at 22:35
    
@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. –  Joe Mar 1 '12 at 22:36
add comment

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...

share|improve this answer
    
You might want to reconsider. –  Gerry Myerson Mar 2 '12 at 0:16
add comment

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

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

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

share|improve this answer
    
Someone care to elaborate the reason for the downvote? –  Mike Mar 1 '12 at 22:45
    
I didn't downvote, but, given the level of the question, I suspect this answer is too elliptical to be very helpful. –  Gerry Myerson Mar 2 '12 at 0:16
    
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. –  Mike Mar 2 '12 at 0:25
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.