The method of substitution in the problem of finding the integral

Question

My teacher gave me a simple problem:
Find $\int \dfrac{x}{\sqrt{x+1}} \, dx$.

This is how I approached it:
I set $u = \sqrt{x+1}$, which implies $u^2 = x + 1$, thus $2u \, du = dx$.

Therefore,
$$ \int \dfrac{x}{\sqrt{x+1}} \, dx = 2 \int \dfrac{u^2 - 1}{u} \, u \, du = 2 \int (u^2 - 1) \, du = \dfrac{2u^3}{3} - 2u + C $$

That will give :
$$ \int \dfrac{x}{\sqrt{x+1}} \, dx = \dfrac{2(\sqrt{x+1})^3}{3} - 2\sqrt{x+1} + C $$

My teacher said it was wrong because $u = \sqrt{x+1}$ cannot be squared and then differentiated.
I don't understand how my approach is incorrect?

Answer 1

There's nothing wrong. Maybe it's more sophisticated than your teacher though. The more traditional approach would probably be to let $u=x+1$ so that $du=dx$ and $$\int\frac{x}{\sqrt{x+1}}dx=\int\frac{u-1}{\sqrt{u}}du=\int(u^{1/2}-u^{-1/2})du=\frac{2}{3}u^{3/2}-2u^{1/2}+C$$ $$=\frac23(x+1)^{3/2}-2(x+1)^{1/2}+C$$ Many different substitutions can work for a given problem.

Answer 2

You could also approach it like this: \begin{align*} \int\frac{x}{\sqrt{x + 1}}\mathrm{d}x & = \int\frac{(x + 1) - 1}{\sqrt{x + 1}}\mathrm{d}x\\\\ & = \int\sqrt{x + 1}\mathrm{d}x - \int\frac{1}{\sqrt{x + 1}}\mathrm{d}x\\\\ & = \int(x + 1)^{1/2}\mathrm{d}x - \int(x + 1)^{-1/2}\mathrm{d}x\\\\ & = \frac{2}{3}(x + 1)^{3/2} - 2(x + 1)^{1/2} + C \end{align*}

Hopefully this helps!

Answer 3

Adding this justification makes your working more rigorous:

• since the result applies to every subinterval of $(-1,\infty),$ it also applies to every subinterval of $[-1,\infty).$

Answer 4

When you use substitution on an integral, you should alway wonder for the definition domain. When you use a linear substitution, the definition domain question is implicit because a linear substitution is a bijective monotonic $C^\infty$ transformation. But as soon as you use square (or any even power) not speaking of periodic functions, you loose the bijective property so the definition domain shall be explicit.

If you fail doing that, you could end with a solution which could be defined outside of the original definition domain, and by definition such a solution cannot be obtained by integrating the original function.

So while your substitution is perfectly correct, using it without the precision $u \in ]0; \infty [$ is not.

Answer 5

My teacher said it was wrong because $u=\sqrt{x+1}$ cannot be squared and then differentiated. I don't understand how my approach is incorrect?

Your approach is fine, and the reason your teacher gave for rejecting it is unsound. You can square $u$ and then differentiate. That's equivalent to performing a second substitution $v = u^{2}$, with $dv = 2udu$. In this case, that gets you to $2udu = dv = dx$. It's all the chain rule, if you want to look at it that way.

There has been some speculation here that your teacher's objection was more about the bounds of $u$ for which the substition is valid, but I'm not inclined to accept that, because

1. that's not what (you say) they actually said, and
2. the more usual substitution $u = x+1$ described in other answers has the same issue.

Perhaps your teacher was tired, or perhaps surprised by your approach. It is a bit more involved and, frankly, insightful than they probably expected. And there certainly are things that you can do with ordinary expressions, but cannot do with differentials. Your teacher has a hard job, so do please cut them some slack.

Answer 6

It's unclear to me what would make a solution to an antidifferentiation question incorrect, other than if the answer's derivative were not the integrand in question.

As far as integration by substitution goes, a sufficient condition that $$ \int_a^b f(g(x)) g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du $$ is that $f$ and $g'$ are both continuous. And there are often many options for the $f$ and $g$ in a particular integration problem. For your problem, two options are:

• $g(x) = x+1$ and $f(u) = \frac{u-1}{\sqrt{u}}$. Then $f(g(x)) = \frac{x}{\sqrt{x+1}}$ and $g'(x) = 1$.
• $g(x) = \sqrt{x+1}$ and $f(u) = 2(u^2-1)$. Then $f(g(x)) = 2x$ and $g'(x) = \frac{1}{2\sqrt{x+1}}$.

In both cases, $f$ and $g'$ are continuous on the domain of the integrand, and $f(g(x))g'(x) = \frac{x}{\sqrt{x+1}}$, as desired.

Starting from the substitution $u=\sqrt{x+1}$, one could say $\frac{du}{dx} = \frac{1}{2\sqrt{x+1}}= \frac{1}{2u}$, so $dx = 2u\,du$ without implicit differentiation. But your method is the same. In practice, integration is a much more formal (meaning, less attention is given to rigor) skill than a lot of single-variable calculus.