Why $\int {dx \over x \sqrt{x+3}} \neq \int {2u\cdot du \over (u^2-3)u} \rvert_{u=\sqrt{x+3}} $?

Question

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!

Answer 1

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,

Answer 2

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.