Thread: Partial fraction with repeated irreducible 2nd-degree polynomial in the denominator

1. Partial fraction with repeated irreducible 2nd-degree polynomial in the denominator

Woe is me and the partial fractions...

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

I tried working with just the fraction and got the same thing back that I put in, then I tried completing the square and subbing u=x+1 and got the same thing .

What the heck do I do with this thing?

2. Originally Posted by symstar
Woe is me and the partial fractions...

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

I tried working with just the fraction and got the same thing back that I put in, then I tried completing the square and subbing u=x+1 and got the same thing .

What the heck do I do with this thing?
$\int\frac{x-3}{(x^2+2x+4)^2}\,dx=\int\frac{x-3}{(x^2+2x+1+3)^2}\,dx=\int\frac{x-3}{((x+1)^2+3)^2}\,dx$

Let $u=x+1\implies \,du=\,dx$

So we get $\int\frac{u-4}{(u^2+3)^2}\,du$

Now we see that we can split it up and get $\int\frac{u\,du}{(u^2+3)^2}-4\int\frac{\,du}{(u^2+3)^2}$

For $\int\frac{u\,du}{(u^2+3)^2}$, let $z=u^2+3\implies \,dz=2u\,du$, then we get $\tfrac{1}{2}\int\frac{\,dz}{z^2}$

For $-4\int\frac{\,du}{(u^2+3)^2}$, let $z=u^2+3\implies \,dz=2u\,du$, then we get $-2\int\frac{u\,dz}{z^2}=-2\int\frac{z-3\,dz}{z^2\sqrt{z-3}}$

let $z=3\sec^2\vartheta\implies \,dz=6\sec^2\vartheta\tan\vartheta\,d\vartheta$

Thus, $-4\int\frac{\,du}{(u^2+3)^2}=-12\int\frac{(3\sec^2\vartheta-3)(\sec^2\vartheta\tan\vartheta)\,d\vartheta}{9\se c^4\vartheta\sqrt{3(\sec^2\vartheta-1)}}=-12\int\frac{3\sec^2\vartheta\tan^3\vartheta\,d\var theta}{9\sqrt{3}\sec^4\vartheta\tan\vartheta}$ $=-\frac{4\sqrt{3}}{9}\int\sin^2\vartheta\,d\vartheta =-\frac{4\sqrt{3}}{9}\left[\tfrac{1}{2}\vartheta-\tfrac{1}{2}\sin\vartheta\cos\vartheta\right]$

Now, converting back to an equation in z, we see that the anti-derivative is $-\frac{4\sqrt{3}}{9}\left[\tfrac{1}{2}\tan^{-1}\left(\frac{\sqrt{z-3}}{\sqrt{3}}\right)-\tfrac{1}{2}\frac{\sqrt{3z-9}}{z}\right]$

The anti-derivative (as a whole) is $-\frac{1}{2z}-\frac{4\sqrt{3}}{9}\left[\tfrac{1}{2}\tan^{-1}\left(\frac{\sqrt{z-3}}{\sqrt{3}}\right)-\tfrac{1}{2}\frac{\sqrt{3z-9}}{z}\right]+C$

Since $z=u^2+3$, then we get $-\frac{1}{2(u^2+3)}-\frac{4\sqrt{3}}{9}\left[\tfrac{1}{2}\tan^{-1}\left(\frac{u\sqrt{3}}{3}\right)-\tfrac{1}{2}\frac{\sqrt{3}u}{u^2+3}\right]+C$

Now, since $u=x+1$ we *finally* see that $\int\frac{(x-3)\,dx}{(x^2+2x+4)^2}=\frac{-1}{2x^2+4x+8}-\tfrac{4\sqrt{3}}{9}\left[\tfrac{1}{2}\tan^{-1}\left(\frac{(x+1)\sqrt{3}}{3}\right)-\tfrac{1}{2}\frac{\sqrt{3}(x+1)}{x^2+2x+4}\right]+C$

I (most likely) made a mistake somewhere in this calculation, but the answer is supposed to be $-\frac{4x+7}{6(x^2+2x+4)}-\frac{2\sqrt{3}}{9}\tan^{-1}\left(\frac{(x+1)\sqrt{3}}{3}\right)+C$

Try to make sense of what I did......

--Chris

3. An easier way to find the integral $\int\frac{du}{(u^2+3)^2}$ is simply to make the substitution $u=\sqrt{3}\tan{z}$. Then $du=\sqrt{3}\sec^2{z}\,dz$, and
$\int\frac{du}{(u^2+3)^2}=\int\frac{\sqrt{3}\sec^2{ z}\,dz}{(3\tan^2{z}+3)^2}$
$\;\;=\frac{1}{3\sqrt{3}}\int\frac{\sec^2{z}}{(\tan ^2{z}+1)^2}\,dz$
$\;\;=\frac{1}{3\sqrt{3}}\int\frac{\sec^2{z}}{(\sec ^2{z})^2}\,dz$
$\;\;=\frac{1}{3\sqrt{3}}\int\frac{1}{\sec^2{z}}\,d z$
$\;\;=\frac{1}{3\sqrt{3}}\int\cos^2{z}\,dz$
Proceeding from there should be much easier.

--Kevin C.

4. Wow. I follow most of it... but, $-2\int\frac{u\,dz}{z^2}=-2\int\frac{z-3\,dz}{z^2\sqrt{z-3}}$ did you rationalize the numerator here?

5. Originally Posted by symstar
Wow. I follow most of it... but, $-2\int\frac{u\,dz}{z^2}=-2\int\frac{z-3\,dz}{z^2\sqrt{z-3}}$ did you rationalize the numerator here?
I'd follow TwistedOne151's way of integrating $-4\int\frac{\,du}{u^2+3}$, but the way I did it was follows:

I saw that if $z=u^2+3$, then $\,dz=2u\,du$

That results in $-2\int\frac{u\,dz}{z^2}$

Now, $z=u^2+3\implies u=\sqrt{z-3}$

So, we now get $-2\int\frac{\sqrt{z-3}\,dz}{z^2}$

Oh...I just noticed now that we could just make the substitution $z=3\sec^2\vartheta$ right here.

But to answer your question, I saw that $\sqrt{z-3}=\frac{z-3}{\sqrt{z-3}}$, so then the integral became $-2\int\frac{(z-3)\,dz}{z^2\sqrt{z-3}}$

As a side note, did anyone catch my little mistake? My answer simplifies to $-\frac{4x+1}{6(x^2+2x+4)}-\frac{2\sqrt{3}}{9}\tan^{-1}\left(\frac{(x+1)\sqrt{3}}{3}\right)$, but the answer is $-\frac{4x+{\color{red}7}}{6(x^2+2x+4)}-\frac{2\sqrt{3}}{9}\tan^{-1}\left(\frac{(x+1)\sqrt{3}}{3}\right)$

I probably shouldn't be doing these problems when I'm dead tired

--Chris

6. Following Chris' will... lol

Originally Posted by Chris L T521
For $-4\int\frac{\,du}{(u^2+3)^2}$, let $z=u^2+3\implies \,dz=2u\,du$, then we get $-2\int\frac{u\,dz}{z^2}=-2\int\frac{z-3\,dz}{z^2\sqrt{z-3}}$
Wow my eyes are all blurry

I think the mistake is here.

In order not to do any mistake, I always write what du is (or at least, I do it to make people understand ): $du=\frac{dz}{2u}$

So $-4 \int \frac{du}{(u^2+3)^2}$ transforms into $-2 \int \frac{dz}{{\color{red}u} z^2}$

I haven't checked what followed..

7. Originally Posted by Moo
Wow my eyes are all blurry

I think the mistake is here.

In order not to do any mistake, I always write what du is (or at least, I do it to make people understand ): $du=\frac{dz}{2u}$

So $-4 \int \frac{du}{(u^2+3)^2}$ transforms into $-2 \int \frac{dz}{{\color{red}u} z^2}$

I haven't checked what followed..
Wow...that was....erm...yea

I think that is the only mistake

--Chris