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

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.

Thanks in advance!


@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.

  • 1
    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.2011-02-16

3 Answers 3

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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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    @Chandru1: Thanks, I see now that I was forgetting the $\sqrt{2}$ was in place for the 2 so when you square it to be x^22011-02-16
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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: Thanks a bunch, that's where I kept getting lost. So when you factor out the $\frac{1}{\sqrt{8}}$ the other $\frac{\sqrt{2}}{\sqrt{8}}$ is already there which is how Chandru1 got the 8 in the denominator of the constant or $\frac{\sqrt{2}}{8}$ rather that was confusing me. I guess it was due to the fact that I was substituting in the integral the wrong which way which is why I was confused about the constant that gets factored out (because I substituted wrong).2011-02-16
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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!