# What Techniques of Integration would be best suited for this Integral?

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.

• Try an initial $u$-substitution as $u=x+1$ and see what happens... Jun 16 '16 at 4:12
• An alternative substitution that is useful is $u = \sqrt{x+1}$. Jun 16 '16 at 4:16
• 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. Jun 16 '16 at 4:47
• 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. Jun 16 '16 at 5:05

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

• That's really clever! Jun 16 '16 at 4:22
• 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. Jun 16 '16 at 4:34

Use the substitution method.

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

• I totally forgot about special substitutions. I will try it right now. Thanks! Jun 16 '16 at 4:20
• No problem, and good luck on your test. Jun 16 '16 at 4:21
• 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! Jun 16 '16 at 4:39

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}

• 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. Jun 16 '16 at 4:37