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$$\int \sqrt{x^2 + 2x}dx$$

I have no clue what to do on this problem. It is in the trig substitution chapter so I know I have to use that somehow. I know that I can not complete the square because both terms are positive and will not give me a difference of squares.

I know u subsitution will not work because I get leftover x terms.

I know that I basically have to manipulate this problem algebraically before I can work with it but I just do not know how to do that.

I tried to factor out an x or -x but neither makes progress.

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Complete the square: $x^2+2x=(x+1)^2-1$. –  David Mitra Jun 9 '12 at 1:48
@Jordan: Since you have asked quite a few question last days, I wanted to make sure that you're aware of 50 questions/month limit, see meta: How much asking is too much? I thought this information might be useful for you. –  Martin Sleziak Jun 9 '12 at 9:28
So it seems that I was wrong about the limit on the number of questions per month, see here: Have the limits on number of questions per month/day been increased (or cancelled)?. –  Martin Sleziak Jul 2 '12 at 11:47
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2 Answers

up vote 4 down vote accepted

Same thing as the last one. Use $$ \int \sqrt{x^2 + 2x }\ \ dx = \int \sqrt{x^2 + 2x + 1 - 1}\ \ dx = \int \sqrt{(x + 1)^2 - 1}\ \ dx $$ and use the hint $\sec \theta = x+1$.

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I think I have been doing math for too long today, this was really obvious and I spent about 20 minutes just trying to complete the square. –  user138246 Jun 9 '12 at 1:55
@Jordan You're definitely improving. Keep it up. –  Eugene Jun 9 '12 at 1:56
$dx$'s tsk tsk. –  Dylan Moreland Jun 9 '12 at 2:47
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Ok. Here is a general method of solving integrals of the form $P(x) = ax^{2}+bx+c$.

\begin{align*} ax^{2}+bx+c &= a \cdot(x^{2}+\frac{b}{a} x) + c \\\ &= a \cdot \biggl(x^{2}+ \frac{b}{a} \cdot x +\frac{b^2}{4a^2}\biggr) + c - \frac{b^2}{4a} \\\ &= a \cdot \biggl(x+\frac{b}{2a}\biggr)^{2} + c-\frac{b^2}{4a} \end{align*}

Now after doing this put $\displaystyle x+\frac{b}{2a} = \frac{1}{\sqrt{a}} \cdot \sqrt{\biggl(c-\frac{b^2}{4a}\biggr)} \cdot \tan\theta$

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Nice answer chandu +1. –  Iyengar Jun 9 '12 at 16:16
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