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When trying to integrate the following, I thought these where equal: $$\int {dx \over x \sqrt{x+3}} = \left.\int {2u\cdot du \over (u^2-3)u} \right|_{u=\sqrt{x+3}} $$

But they are not. If you set $x=1$, you get $\int {dx \over 2}$ and $\int2du$ respectively. As far as I can see my u-sub is valid, but I don't understand why I get the constants ${1 \over 2}$ and $2$. I would have thought they'd be the same. Or are they different becuase of the substitution. It could be I've been staring at this problem way to long after a long day of homework. So If I'm missing something really simple in my understand please call me out on it!

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You don't ''set $x=1$'' in an integral. That is not a valid operation. It would be just as strange as consider the sum $$ \sum_{i=0}^n a_i $$ and setting "$i=2$" to say you get $\sum_{i=0}^n a_2$. It makes no sense to do that ; the $x$ in $\int f(x) \, dx$ is analogous to the $i$ in a summation ranging with $i$ from $0$ to $n$. The part which is not analogous is that you can integrate "without bounds" (i.e. you can find primitives) which makes little sense while computing discrete sums. But with integration bounds, the analogy working.

Also, if your substitution was successful, you should end up with $du$ in the second integral, not $dx$.

Finally, if you put $u = \sqrt{x+3}$, you have $du = \frac 1{2 \sqrt{x+3}} dx$, so $2 du = \frac{dx}{\sqrt{x+3}}$ and $\frac 1x = \frac 1{u^2-3}$. Therefore $$ \int \frac{dx}{x\sqrt{x+3}} \underset{u = \sqrt{x+3}}{\longrightarrow} \int \frac{2 \, du}{u^2-3}. $$ Hope that helps,

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  • $\begingroup$ The whole reason I was trying to see if the integrals were the same was because I took the u-sub step that I did and Wolfram Alpha u-sub'd on $u = x+3$. It also got a different answer than I did in the end. So I wanted to see if the function being integrated was the same (i.e. did I make a mistake doing my u-sub or algebra). I now get that setting x equal to one doesn't do what I wanted it to do for checking for this. But nonetheless when I run both of the integrals thru Wolfram Alpha I get different answers. I'm confused as to why. (NOTE: I had $dx$ in the 2nd integral as a typo). $\endgroup$ – Matt Sep 21 '14 at 1:26
  • $\begingroup$ @Matt : Did you notice my result for the substitution $u = \sqrt{x+3}$ is not the same as yours? $\endgroup$ – Patrick Da Silva Sep 21 '14 at 1:46
  • $\begingroup$ P.S. : You wrote $u = x+3$ in your comment... I assume you meant $u = \sqrt{x+3}$, but just in case, the substitution $u=x+3$ would've given you the integral $$ \int \frac{du}{(u-3)\sqrt u} $$ $\endgroup$ – Patrick Da Silva Sep 21 '14 at 1:47
  • $\begingroup$ The result looks the same to me (you just canceled out a u in the denominator and numerator). Yes, I meant $u=x+3$. That's what Wolfram Alpha uses on the first integral. Apparently, the two integrals when evaluated on that website are equal dispute very different looking answers(one involves logs and radicals, the other uses $arctanh$ and no logs). The difference thru me off (and is what led me to the invalid idea of setting $x=1$ on an integral to try to see why they seemingly weren't equal). $\endgroup$ – Matt Sep 21 '14 at 1:54
  • $\begingroup$ @Matt : Oh, sorry I didn't see that $u$ in the denominator (why is it there in the first place?... :( ) The thing we keep repeating students during integration courses is that using computers to verify indefinite integrals can be confusing ; the results will all be equal up to a constant, and sometimes, up to a constant, functions can have a veeeeeeeery different appeareance. $\endgroup$ – Patrick Da Silva Sep 21 '14 at 2:23
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The substitution is valid. You cannot expect to get the same value, since you are adding a factor of $2u$. You can try it on any simple substitution and you'll see the result probably won't be the same.

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