Integration by Partial Frac.

Question

Where do I go on this integration by partial fractions problem?

https://i.stack.imgur.com/CkEjN.jpg

Answer 1

We have

$f(x)=\frac{x^2}{x^4-2x^2-8}=$

$\frac{x^2}{(x-2)(x+2)(x^2+2)}=$

$\frac{a}{x-2}+\frac{b}{x+2}+\frac{cx+d}{x^2+2}$

this gives

$a=\frac{1}{6}$

obtained by multiplying by $(x-2)$ and replacing $x$ by $2$.

$b=\frac{-1}{6}$

obtained by multiplying by $(x+2)$ and replacing $x$ by $-2$.

$c=-a-b=0$

obtained by multiplying by $x$ and $x\to \infty$

and

$d=a-b=\frac{1}{3}$

obtained by replacing $x$ by $0$.

thus

$f(x)=\frac{1}{6}(\frac{1}{x-2}-\frac{1}{x +2}+\frac{1}{1+(\frac{x}{2})^2})$

and the primitive

$F(x)=\frac{1}{6}(ln( |\frac{x-2}{x+2}|$

$+2arctan(\frac{x}{2} ))+C$

defined in $(-\infty,-2) $ or $(-3,2)$ or $(2,+\infty)$.

Answer 2

You can either substitue values in for $x$ like $2$ that cancel out other terms or you can actually foil it all out and then set set up a system of equations with the coefficients of terms of equal degree. For instance, if $x^2+2x=Ax^2+Ax+Bx$, you would have $Ax^2=x^2\implies A=1$ and $Ax+Bx=2x\implies A+B=2$.

Answer 3

HINT:

Put $x^2=y$ to get $$\dfrac y{y^2-2y-8}=\dfrac y{(y-4)(y+2)}=\dfrac A{y-4}+\dfrac B{y+2}$$