# Help solving another integral $\int (2x^2+4x-2)^{-\frac{3}{2}} \ dx$

$$\int (2x^2+4x-2)^{-\frac{3}{2}} \ dx$$

Complete the square?

$$\int \frac{1}{(2(x+1)^2-4)^\frac{3}{2}} \ dx$$

Not sure what do do next, or if I should try something else?

Big help if you can show as "step-by-step" possible as you can.

@Chandru1:

I'm confused at what happens to the 2 that gets factored out when completing the square (the 2 in the denominator) because it looks like Arturo left it out and instead substituted:

$\int \frac {1}{(2(x+1)^2-4)^\frac{3}{2}} \ dx$

$u = x+1$ $du =dx$

which changes the integral to $\int \frac {1}{(2(u^2-4))^\frac{3}{2}}$

Would I be best factoring it out as a 1/2 in front of the integral or leaving it in and then using trig. substitution like this?

$u= \sqrt{2}sec(t)$ $du = \sqrt{2}sec(t)*tan(t) dt$

Which changes it to:

$\int \frac {\sqrt{2}sec(t)*tan(t)}{(2(2sec^2(t)-4))^\frac{3}{2}} \ dt$ = $\int \frac {\sqrt{2}sec(t)*tan(t)}{(2(tan^2(t)))^\frac{3}{2}} \ dt$ = $\int \frac {\sqrt{2}sec(t)*tan(t)}{(2tan^3(t))} \ dt$

My $\sqrt{2}$ factors out like yours, but I have no 1/8 so figure I factor it out as a 1/2 earlier like I thought? But then the constant in front of my integral would be $\frac{\sqrt{2}}{2}$ and you have $\frac{\sqrt{2}}{8}$

I know what's wrong, just not how I went wrong.

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The second integral is thoroughly messed up; what happened to the $\frac{3}{2}$ exponent? And the "completing" of the square is incorrect as well. – Arturo Magidin Feb 16 '11 at 5:08
... not to mention the unbalanced parentheses... – Arturo Magidin Feb 16 '11 at 5:21
Thanks, I will fix it. I'm sorry, I'm still getting used to using the TEX. – Finzz Feb 16 '11 at 8:31
Yes, I left the $2^{-3/2}$ that I factored out. I said "factor it out", and then I said the integral you want was up to a constant the integral I give. If you do the integral I did, at the end you would have to multiply the answer by $2^{-3/2}$. I just saw no point in carrying around the constant when I expected the trouble was with the integral. But clearly not, so I'll add it back in. – Arturo Magidin Feb 16 '11 at 15:36

Hint. $(2x^{2}+4x-2) = 2(x^{2}+2x +1) -4 = 2(x+1)^{2}-4$. Now subsitute $x+1 = \sqrt{2} \sec{t}$. Then you have $dx = \sqrt{2} \sec{t} \cdot \tan{t} \ dt$. Now, $$2 \cdot \Bigl[(x+1)^{2} -2 \Bigr]= 2 \cdot \Bigl[ 2 \cdot (\sec^{2}{t}-1)\Bigr] = 4 \cdot (\sec^{2}{t}-1)$$.

Now what will you get if you power them by $\frac{3}{2}$. You will get $4^{3/2} \cdot \Bigl[ \tan^{2}{t}\Bigr]^{3/2} = 8 \cdot \tan^{3}{t}$.

And try to evaluate the integral. Simplyfying you should get

\begin{align*} \int \frac{1}{(2x^{2}+4x-2)^{-\frac{3}{2}}} \ \textrm{dx} &= \frac{\sqrt{2}}{8} \int\frac{\sec{t} \cdot \tan{t}}{\tan^{3}{t}} \ \textrm{dt} \\ &= \frac{\sqrt{2}}{8} \int\frac{\cos{t}}{\sin^{2}{t}} \ \textrm{dt} \end{align*}

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I see the sqrt 2 was pulled out, but where did the 1/8 come from? – Finzz Feb 16 '11 at 9:09
@Finzz: Whats $2^{3/2}$. It comes out while subsituting in the denominator. If you don't understand tell me, i shall edit the answer. – anonymous Feb 16 '11 at 9:15
@Chandru1: I will just edit my original question, too long to explain in comments. – Finzz Feb 16 '11 at 9:50
@Chandru1: It would make sense if the radical was the whole 2/8, but you put it just over the 2. That's what I want to know, read the second half of my question to see what I mean. You ask what is 2^(3/2), well its $\sqrt{8}$, but if you look at arturo – Finzz Feb 16 '11 at 10:52
Got cut off on that comment:--------------------------------- You ask what is 2^(3/2), well its $\sqrt{8}$, but if you look at Arturo's last integral, you would factor it out and have a constant of $\frac{\sqrt{2}}{\sqrt{8}}$ not what you have. – Finzz Feb 16 '11 at 10:58

First, factor out a $2$: $$(2x^2 + 4x - 2) = 2(x^2+2x-1)\quad\text{so}\quad (2x^2+4x-2)^{-3/2} = 2^{-3/2}(x^2+2x-1)^{-3/2}.$$ Do this because you can pull it out of the integral and it will make things easier.

So, up to a constant, this is the same as doing the integral $$\frac{1}{2^{3/2}}\int\frac{dx}{(x^2+2x-1)^{3/2}}.$$ Completing the square is a good first step. $x^2+2x-1 = (x^2+2x+1) - 2 = (x+1)^2 - 2$. So we get the integral $$\frac{1}{2^{3/2}}\int\frac{dx}{\bigl( (x+1)^2 - 2\bigr)^{3/2}}.$$ Doing a change of variable $u=x+1$ changes it to $$\frac{1}{2^{3/2}}\int\frac{du}{(u^2-2)^{3/2}}.$$

Now try a trigonometric substitution to get rid of that pesky square root in the exponent. Use $\tan^2\theta + 1 = \sec^2\theta$, or $\sec^2\theta - 1 =\tan^2\theta$. So set $u = \sqrt{2}\sec\theta$ to get $$u^2-2 = 2\sec^2\theta - 2 = 2(\sec^2\theta - 1) = 2\tan^2\theta$$ and $du = \sqrt{2}\sec\theta\tan\theta\,d\theta$, so $$\frac{1}{2^{3/2}}\int\frac{du}{(u^2-2)^{3/2}} = \frac{1}{2^{3/2}}\int\frac{\sqrt{2}\sec\theta\tan\theta\,d\theta}{(2\tan^2\theta)^{3/2}} = \frac{1}{2^{3/2}}\int\frac{\sqrt{2}\sec\theta\tan\theta\,d\theta}{2^{3/2}\tan^3\theta}.$$ Now work the trigonometric integral.

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@Arturo: Thats what i precisely did. – anonymous Feb 16 '11 at 5:18
@Chandru1: Goody for you. – Arturo Magidin Feb 16 '11 at 5:19
@Arturo: Shouldn't the second integral have the $2^\frac{-3}{2}$ in front of it? I'm confused at what happens to the 2 that you factored out in the beginning. – Finzz Feb 16 '11 at 8:39
Nevermind, saw you stuck it back in at the end, jumped the gun. – Finzz Feb 16 '11 at 8:48
@Finzz: I took out the $2^{-3/2}$ at the beginning, and then said I would only do the integral that was left. (Notice I said, "up to a constant, this is the same as doing the integral...") The rest of the post is just about the integral that is left, namely $\int(x^2 + 2x - 1)^{-3/2}dx$. So, yes, you are supposed to be "carrying" the $2^{-3/2}$ around, but I expected that the trouble was with the integral, not the extra factor of $2^{-3/2}$ that is only multiplying the whole thing. – Arturo Magidin Feb 16 '11 at 15:33

Can you see that the expression in the parentheses can be written as $((x+1)^2-2)$. Do we see a difference of squares here?

Next a u sub is required. $U=(x+1)-\sqrt 2$

Fill in the details.

A partial fraction decomp of $1/((u+2\sqrt 2)u^2)$ will be needed.

And the solution has become as sweet as apple pie!

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