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I'm trying to solve this simple integral:

$$\frac12 \int \frac{x^2}{\sqrt{x + 1}} dx$$

Here's what I have done so far:

  1. $\displaystyle t = \sqrt{x + 1} \Leftrightarrow x = t^2 - 1 \Rightarrow dx = 2t dt$

  2. $\displaystyle \frac12 \int \frac{x^2}{\sqrt{x + 1}} dx = \int \frac{t (t^2 - 1)^2}t dt$

  3. $\displaystyle \int (t^2 - 1)^2 dt = \frac15 t^5 - \frac23 t^3 + t + C$

  4. $\displaystyle \frac15 t^5 - \frac23 t^3 + t + C = \frac15 \sqrt{(x + 1)^5} - \frac23 \sqrt{(x + 1)^3} + \sqrt{x + 1} + C$

WolframAlpha tells me steps 1 and 3 are right so the mistake must be somewhere in steps 2 and 4, but I really can't see it.

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This looks fine. I imagine WA further simplified. For example $\sqrt{(x+1)^5}=(x+1)^2\sqrt{x+1}$. –  David Mitra Feb 5 '13 at 17:46
    
@DavidMitra: Actually I checked computing the derivative of my result in WA, and it was giving a long expression. However I didn't notice it was supposing that x was complex. –  hey hey Feb 5 '13 at 17:50
    
Your substitution is good, I like to get rid of square roots. In this case, it might have been marginally easier to let $u=x+1$, so $x^2=(u-1)^2$, and we are integrating $u^{3/2}-2u^{1/2}+u^{-1/2}$. –  André Nicolas Feb 5 '13 at 17:53
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3 Answers

up vote 2 down vote accepted

No, everything is fine. Now simplify what you have got after integration. Note that $\sqrt{a^5}=\sqrt{a}.a^2$ etc.

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There is a (slightly) more obvious way of solving it: rewrite the numerator as $x^2+1-1$ and then the whole integral as a sum of two integrals: $$ \int \frac{(x^2-1)dx}{\sqrt{x+1}} + \int \frac{dx}{\sqrt{x+1}} $$ The second integral is easy, the first one is $$ \int \frac{(x^2-1)dx}{\sqrt{x+1}} =\int \frac{(x+1)(x-1)dx}{\sqrt{x+1}}=\int x \sqrt{x+1}dx-\int \sqrt{x+1}dx $$ The second integral is also easy, and the integrand in the first one should be rewritten and $(x+1-1)\sqrt{x+1}$ and the rest is easy. This is a bit too straightforward, I admit.

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$$\begin{align}\frac{x^2}{\sqrt{x+1}} &= \frac{x^2-1}{\sqrt{x+1}} + \frac{1}{\sqrt{x+1}}\\&=(x-1)\sqrt{x+1} + \frac{1}{\sqrt{x+1}}\\&=(x+1)^{3/2} - 2\sqrt{x+1} + \frac{1}{\sqrt{x+1}}\end{align}$$

I think you can integrate each of these terms.

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