How Do You Solve Integrals Using Partial Fractions?

  • Thread starter Thread starter zeppelinpage4
  • Start date Start date
  • Tags Tags
    Fraction Partial
zeppelinpage4
Messages
1
Reaction score
0

Homework Statement



I basically need to find the integral of (-5x^3-1x^2+4)/(x^4+2x^3)


Homework Equations



Just the basic rules associated with partial fractions when re-writing a polynomial with constants.

The Attempt at a Solution



Since the degree of the denominator is 4 and the degree of the numerator is 3 I went straight to factoring out the denominator.
I ended up getting
(x^4+2x^3)=x^3(x+2)

From this point I went forward with the basic steps (I think I made the mistake here, I'm not sure how to re-write the form for (x^3), so I broke it into (x^2) & (x)).

[(-5x^3-1x^2+4)/(x^4+2x^3)]= [(Ax+B)/(x^2)]+[C/x]+[D/(x+2)]

When I multiply both sides by (x^4+2x^3) (this is the denominator on the left side)

I end up with

(-5x^3-1x^2+4)=x^3(A+C+D) + x^2(2A+B+2C) + x(2B)

Normally i'd solve for the constants A, B, C and D to fit the polynomial on the left side BUT I don't know how to account for the "+ 4" at the end of (-5x^3-1x^2+4).

I know
A+C+D=-5
2A+B+2C=-1
2B=0

?=4

I'm very lost at how to approach this and any advice or help that could give me some clue would be very much appreciated. It's been a tough week with midterms and my calc teacher hasn't been helping. =P
 
Last edited:
Physics news on Phys.org
This bit (x^4+2x^3)=x^3(x+2) is fine.

Personally I would look at the last two terms of the numerator, ask if they look anything familar. I find using that it falls out quite simply to something you can integrate. When you have got that, knowing the answer you will be able to check through your general method, which is what you have to use when you are not so lucky, and clear up where you went wrong maybe.

But there is virtue in simplicity, and I wouldn't present an unnecessarily laborious version.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top