Partial Fraction: Solving x^3+1/x^2+4

  • Thread starter Thread starter tnutty
  • Start date Start date
  • Tags Tags
    Fraction Partial
tnutty
Messages
324
Reaction score
1

Homework Statement


x^3+1 / x^2+4


Homework Equations




The Attempt at a Solution



after using long division i get : x + -(4x-1) / (x^2+4)

then using partial fraction :

-4x-1 / x^2+4 = bx+c / x^2+4

then i am stuck because if what i have
is correct then they both are the same and in the same form.
 
Physics news on Phys.org
tnutty said:
-4x-1 / x^2+4 = bx+c / x^2+4

then i am stuck because if what i have
is correct then they both are the same and in the same form.

-\frac{4x-1}{x^2+4}=\frac{1}{x^2+4}-\frac{4x}{x^2+4}...:wink:
 
Once you gotten to (-4x-1)/(x^2+4) you probably want to leave the partial fractions trail. Split it into -4x/(x^2+4) and -1/(x^2+4). The first one is a simple substitution. I'm going to guess the second one might be something related to arctan(x)? Try a trig substitution.
 
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