0
$\begingroup$

I've been doing integrals non stop preparing for my exam tomorrow, and one has left me stumped for a few days. I've tried coming back to it several times, but I can't seem to manipulate it. It is as follows:

$$\int\frac{x^2}{\sqrt{x+1}}\;dx$$

I've tried different u-sub methods but nothing seems to work for me. My next thought was integration by parts, but when I attempt that method my answer isn't even close to the answer key given by my professor. The answer key shows this as the answer:

$$\frac25(x+1)^\frac{5}{2}-\frac43(x+1)^\frac{3}{2}+2\sqrt{x+1}+C$$

What would be the best technique to handle this integral? I feel like i'm missing something simple, but after 2 days it still hasn't come to me and it's quite frustrating. Thanks in advance.

$\endgroup$
4
  • $\begingroup$ Try an initial $u$-substitution as $u=x+1$ and see what happens... $\endgroup$
    – abiessu
    Jun 16 '16 at 4:12
  • $\begingroup$ An alternative substitution that is useful is $u = \sqrt{x+1}$. $\endgroup$
    – okrzysik
    Jun 16 '16 at 4:16
  • $\begingroup$ Whenever I can, my first reaction is always to try to get rid of radicals. So, as already commented and answered, $u = \sqrt{x+1}$ would be my first (and last) choice. $\endgroup$ Jun 16 '16 at 4:47
  • $\begingroup$ You prefer this over letting $u = x+1$? To me this seems easier since $du=dx$. I'm genuinely curious though as I want to diversify my "calculus toolbox" as much as I can. $\endgroup$ Jun 16 '16 at 5:05
4
$\begingroup$

Let me try: $$\frac{x^2}{\sqrt{x+1}} = \frac{(x+1)^2 - 2(x+1) + 1}{\sqrt{x+1}} = (x+1)^{\frac{3}{2}} -2 (x+1)^{\frac{1}{2}} + (x+1)^{\frac{-1}{2}}$$

$\endgroup$
2
  • $\begingroup$ That's really clever! $\endgroup$ Jun 16 '16 at 4:22
  • $\begingroup$ Indeed it is. It was hard for me to visualize at first, but once I made the recommended substitution it became clear as day! Interesting way to simplify. $\endgroup$ Jun 16 '16 at 4:34
2
$\begingroup$

Use the substitution method.

Let $u=x+1$, then $u-1=x$ and $du=dx$. Try it.

$\endgroup$
3
  • $\begingroup$ I totally forgot about special substitutions. I will try it right now. Thanks! $\endgroup$ Jun 16 '16 at 4:20
  • 1
    $\begingroup$ No problem, and good luck on your test. $\endgroup$ Jun 16 '16 at 4:21
  • 2
    $\begingroup$ Your hint was the easiest for me to understand and I was able to reach the solution easily using your substitution method. Thank you very much for your input, this was a u-sub case I had totally forgotten about! $\endgroup$ Jun 16 '16 at 4:39
1
$\begingroup$

You should substitute $x+1=u^2$ then the rest will become straight forward.

\begin{align} &I=\int \!{\frac {{x}^{2}}{\sqrt {x+1}}}\,{\rm d}x\\ &=\int (\!2\,{u}^{4}-4\,{u}^{2}+2\,{\rm d})u\\ &=\int \!2\,{u}^{4}\,{\rm d}u+\int \!-4\,{u}^{2}\,{\rm d}u+\int \!2 \,{\rm d}u\\ &=\frac{2}{5}\,{u}^{5}+\int \!-4\,{u}^{2}\,{\rm d}u+\int \!2\,{\rm d}u\\ &=\frac{2}{5}\,{u}^{5}-\frac{4}{3}\,{u}^{3}+\int \!2\,{\rm d}u\\ &=\frac{2}{5}\,{u}^{5}-\frac{4}{3}\,{u}^{3}+2\,u+C\\ &=\frac{2}{5}\, \left( x+1 \right) ^{5/2}-\frac{4}{3}\, \left( x+1 \right) ^{3/2}+2\, \sqrt {x+1}+C \end{align}

$\endgroup$
1
  • 1
    $\begingroup$ Thank you for the detailed response. This isn't the way I would immediately think to do things, but it's always nice to see things from someone else's perspective :) I went the $u=x+1, x=u-1$ route instead which worked the same. $\endgroup$ Jun 16 '16 at 4:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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