Question

I am having a ton of difficulty seeing how to proceed when solving this partial fraction integration problem. Anyone have some tips/tricks when dealing with something like this??The integral of : (x^2-3x+7)/(x^2-4x+6)^2

Answer 1

ick

Answer 2

\[\frac{Ax+B}{x^2-4x+6}+\frac{Cx+D}{(x^2-4x+6)^2}\]

Answer 3

First check if the polynomial in the denominator has real roots:
\[\text{discriminant}=(-4)^2-4(1)(6)=16-24<0~~\implies~~\text{no real roots}\]
which means the denominator is irreducible into linear terms. You'll have to settle for the form provided above.

Answer 4

but it has been cooked up to give nice whole number solutions
http://www.wolframalpha.com/input/?i=partial+fractions+%28x^2-3x%2B7%29%2F%28x^2-4x%2B6%29^2

Answer 5

Yeah, I've seen the answer. But i'm trying to understand their logic and see how to proceed were I to get another beautiful question like this. Thanks for the help :)

Answer 6

\[\begin{align*}
\frac{x^2-3x+7}{(x^2-4x+6)^2}&=\frac{Ax+B}{x^2-4x+6}+\frac{Cx+D}{(x^2-4x+6)^2}\\\\
x^2-3x+7&=(Ax+B)(x^2-4x+6)+Cx+D\\\\
&=Ax^3+(-4A+B)x^2+(6A-4B+C)x+(6B+D)
\end{align*}\]
Matching up coefficients gives the system
\[\begin{cases}A=0\\-4A+B=1\\6A-4B+C=-3\\6B+D=7\end{cases}\]

Answer 7

Thanks a ton Sith! Thats beautiful. I've just completed that and i've integrated the first function \[1/(x^2-4x+6) \to \int\limits 1/((x-2)^2+2 ) \to (1/\sqrt{2})\arctan(x/\sqrt{2})\]

Answer 8

currently working on \[\int\limits (x+1)/(x^2-4x+6)dx\]

Answer 9

In your first integral, you're missing a small detail.
\[\int\frac{du}{u^2+a^2}=\frac{1}{a}\arctan\frac{u}{a}\]
but here \(u=x-2\), not \(x\).

Answer 10

Thank you! Yes, and that is how I manage to lose tons of marks in exams....

Answer 11

Np :) What are you trying for the second integral?

Answer 12

I want to find a value for which the derivative is x+1 so I can cancel the numerator. Does that seem feasible?

Answer 13

It is, but it's not the only way to approach it. In the end you'll have to use a trig sub anyway.

Answer 14

Are there more efficient methods that you would recommend?

Answer 15

Not for an indefinite integral, no. I'd go straight for a trig sub. Complete the square again, etc.

Answer 16

Through manipulation of the numerator i think i've got it. If at all possible can you confirm this makes sense. \[\int\limits(x+1)/(x^2-4x+6)dx \to \int\limits (x-2+3)/(x^2-4x+6)dx\]
through subsitution
\[u=x^2-4x+6 \to du=2x-4dx \to 1/2du=x-2dx\]
\[1/2\int\limits3du/u \to 3/2\int\limits du/u \to 3/2\ln \left| u \right| \to 3/2\ln \left| x^2-4x+6 \right|+C\]

Answer 17

Keep in mind, the denominator is being squared!

Answer 18

You just broke my heart. You're right again... I think I need a breather... making so many stupid mistakes.
Alright I'll get back at it. You've been great. :)

Answer 19

So just to confirm though, my manipulating of the numerator and everything I've done omitting the fact that the denom is squared was correct and not something "illegal".

Answer 20

Right, your approach just uses the fact that \(a-a=0\) for all real \(a\). So \(b+a-a=b\) for all real \(b\).

Answer 21

It seems there is a mistake nonetheless. I cannot assume that it will become 3du/u. The 3 should become its own integral since the three isn't a coefficient. As many mistakes as I am making, I am learning. :) Fall down, get back up again lol. Loads of falling down tonight.

Answer 22

Thanks for the help SithsAndGiggles. I'll be moving on to the next question.

Answer 23

Here's how I would have worked the squared-denominator integral:
\[\int\frac{x+1}{(x^2-4x+6)^2}\,dx=\int\frac{x+1}{((x-2)^2+2)^2}\,dx\]
Substitute \(\dfrac{x-2}{\sqrt2}=\tan t\), so \(x=\sqrt2\tan t+2\) and \(dx=\sec^2t\,dt\).
\[\begin{align*}\int\frac{x+1}{((x-2)^2+2)^2}\,dx&=\int\frac{\sqrt2\tan t+3}{(2\tan^2t+2)^2}\sec^2t\,dt\\\\
&=\frac{1}{4}\int\frac{\sqrt2\tan t+3}{(\sec^2t)^2}\sec^2t\,dt\\\\
&=\frac{1}{4}\int\frac{\sqrt2\tan t+3}{\sec^2t}\,dt\\\\
&=\frac{\sqrt2}{4}\int \sin t\cos t\,dt+\frac{3}{4}\int \cos^2t\,dt\\\\
&=\frac{\sqrt2}{2}\int \sin2t\,dt+\frac{3}{8}\int (1+\cos2t)\,dt
\end{align*}\]

Answer 24

If I were to use your approach:
\[\int\frac{x+1}{(x^2-4x+6)^2}\,dx=\frac{1}{2}\int\frac{2x-4}{(x^2-4x+6)^2}\,dx+\int\frac{3}{((x-2)^2+2)^2}\,dx\]
Sub for the denominator in the first integral, and the previously used trig sub for the rightmost integral. \(y=x^2-4x+6\) so that \(dy=(2x-4)\,dx\); \(x-2=\sqrt2\tan z\), so \(dx=\sqrt 2\sec^2z\,dz\).
\[\begin{align*}\int\frac{x+1}{(x^2-4x+6)^2}\,dx&=\frac{1}{2}\int\frac{dy}{y}+\frac{3}{4}\int\frac{\sec^2z}{(\tan^2z+1)^2}\,dz\\\\
&=\frac{1}{2}\int\frac{dy}{y}+\frac{3}{4}\int\cos^2z\,dz\\\\
&=\frac{1}{2}\int\frac{dy}{y}+\frac{3}{8}\int(1+\cos2z)\,dz
\end{align*}\]

Answer 25

Oops, missing a factor of \(\sqrt2\) in the second numerator, but you get the idea.

Answer 26

Also, should be \(\dfrac{dy}{y^2}\) >_<

Answer 27

Yeah. I'm sure they probably wanted us to use that considering we have just finished trig sub. The factor you're missing comes from your \[x=2√tant+2 \] you forgot the \[sqrt{(2)}\] in front of your sec. \[dx=sec2tdt.\]

Answer 28

lol sorry about the format there....

Answer 29

I find these questions hard due to the incredible amount of areas where you can make a silly mistake. :)

Answer 30

THey are fun challenges though.

Answer 31

Integrating rational functions via partial fractions can get pretty tedious, and it doesn't help that there's so much room to make mistakes...