Partial fractions (1 Viewer)

asianese · May 31, 2012

The usual procedure of partial fractions is simple, but I am having trouble with fractions such as $\frac{1+x^2}{1+x^4}$ and perhaps $\frac{1}{1+x^4}$ ? How is the denominator going to become a series of factors??

Thanks.

fortune_cookie · May 31, 2012

You could use a completion of the square method

add and subtract 2x^2 on the bottom, then factor as a difference of two sqaures. Only problem is that results in radicals. Gets bit messy.

fortune_cookie · May 31, 2012

1/[ (x^2+1)^2 - (sqrt(2) x)^2 ] = 1/[ (x^2 + sqrt(2)x + 1) (x^2 -sqrt(2)x +1) ]

etc

Rewrite as (Ax+B)/[ (x^2 + sqrt(2)x + 1)] +(Cx+D)/[ (x^2 -sqrt(2)x +1)]

asianese · May 31, 2012

Ah...I see! Thankyou.

SpiralFlex · May 31, 2012

$I strongly am against partial fractions in this case as it is very very tedious. Alternatively, you can use the results of,$

[HR][/HR]

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

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

$\int \frac{1+x^2}{1+x^4}\;dx$

$=\int \frac{\frac{1}{x^2}+1}{\frac{1}{x^2}+x^2}\;dx$

$=\int \frac{\frac{1}{x^2}+1}{x^2+\frac{1}{x^2}-2+2}\;dx$

$=\int \frac{\frac{1}{x^2}+1}{(x-\frac{1}{x})^2+2}\;dx$

$$Let$\; u=x-\frac{1}{x}$

$du=\frac{1}{x^2}+1 \;dx$

$=\int \frac{du}{u^2+2}$

$=\frac{1}{\sqrt{2}}\tan^{-1}\frac{u}{\sqrt{2}}$

$=\frac{1}{\sqrt{2}}\tan^{-1}\frac{x-\frac{1}{x}}{\sqrt{2}}+C$

[HR][/HR]

$\int \frac{1}{1+x^4} \;dx$

$=\frac{1}{2}\int \frac{2}{1+x^4} \;dx$

$=\frac{1}{2} \int \frac{(x^2+1)-(x^2-1)}{1+x^4} \;dx$

$=\frac{1}{2}\int \frac{x^2+1}{1+x^4}-\frac{x^2-1}{1+x^4}\;dx$

$Similar process as before,$

$=\frac{1}{2}\int \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}-\frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}}\;dx$

$=\frac{1}{2}\int \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}-2+2}-\frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}+2-2}\;dx$

$=\frac{1}{2}\int \frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})^2+2}-\frac{1-\frac{1}{x^2}}{(x+\frac{1}{2})^2-2}\;dx$

$For the first integral,$

$$Let$\; t=x-\frac{1}{x}$

$dt=1+\frac{1}{x^2} \;dx$

$For the second integral,$

$$Let$\; u=x+\frac{1}{x}$

$du=1-\frac{1}{x} \;dx$

$Finish it off with the same process,$

$\frac{1}{2\sqrt{2}}(\tan^{-1} \frac{x-\frac{1}{x}}{\sqrt{2}}+\frac{1}{2}\ln |\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}|)+C$

Make a Difference – Donate to Bored of Studies!

Partial fractions (1 Viewer)

asianese

Σ

fortune_cookie

Banned

fortune_cookie

Banned

asianese

Σ

SpiralFlex

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)