partial fractions problem

**Abbas** · December 30th 2009, 12:37 PM

Hi all ...

Here is an integration which we use partial fractions
I coulnd get the four constants .. any help?

$\int \frac{4x^2+3}{(x^2+x+1)^2}dx =$

**pickslides** · December 30th 2009, 01:37 PM

$ x^2+x+1 = \left(x+\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)\lef t(x+\frac{1}{2}-\frac{\sqrt{3}}{2}i\right) $

**Soroban** · December 30th 2009, 03:24 PM

Hello, Abbas!

Here is an integration which we use partial fractions.
I couldn't get the four constants .. any help?

$\int \frac{4x^2+3}{(x^2+x+1)^2}\,dx$

We have: . $\frac{4x^2 + 3}{(x^2+x+1+1)^2} \;=\;\frac{Ax}{x^2+x+1} + \frac{B}{x^2+x+1} + \frac{Cx}{(x^2+x+1)^2} + \frac{D}{(x^2+x+1)^2}$

Then: . $4x^2+3 \;=\;Ax(x^2+x+1) + B(x^2+x+1) + Cx + D$

$x =1:\quad \;\:7 \:=\: A(1)(3) + B(3) + C(1) + D$

$x = \text{-}1: \quad \text{-}1 \:=\: A(-1)(1) + B(1) + C(-1) + D$

$x=2:\quad\, 35 \:=\: A(2)(7) + B(7) + C(2) + D$

$x=0:\quad\;\: 3 \:=\: A(0) + B(1) + C(0) + D$

Solve the system: . $\begin{array}{ccc}3A + 3B + C + D &=& 7 \\ \text{-}A + B - C + D &=& \text{-}1 \\ 14A + 7B + 2C + D &=& 36 \\ B + D &=& 3 \end{array}$

You should get: . $\begin{Bmatrix}A &=& 4 \\ B &=&\text{-}4 \\ C &=& 0 \\ D &=& 7 \end{Bmatrix}$

**skeeter** · December 30th 2009, 04:04 PM

. $\frac{4x^2 + 3}{(x^2+x+1+1)^2} \;=\;\frac{Ax}{x^2+x+1} + \frac{B}{x^2+x+1} + \frac{Cx}{(x^2+x+1)^2} + \frac{D}{(x^2+x+1)^2}$

$4x^2+3 \;=\;Ax(x^2+x+1) + B(x^2+x+1) + Cx + D$

$4x^2 + 3 = Ax^3 + (A+B)x^2 + (A+B+C)x + (B+D)$

equating coefficients ...

$A = 0$

$B = 4$

$C = -4$

$D = -1$

**Krizalid** · December 30th 2009, 06:15 PM

Originally Posted by Abbas

Hi all ...

Here is an integration which we use partial fractions
I coulnd get the four constants .. any help?

$\int \frac{4x^2+3}{(x^2+x+1)^2}dx =$

I'll compute $\int{\frac{x^{2}}{\left( x^{2}+x+1 \right)^{2}}\,dx}.$ (Note that when computing these you'll realize how to do the other integral, they're the same.)

We have $t^{2}+t+1=\frac{4t^{2}+4t+4}{4}=\frac{(2t+1)^{2}+3 }{4}.$ Now in your integral put $x=\frac1t$ and get $-\int{\frac{dt}{\left( t^{2}+t+1 \right)^{2}}}=-16\int{\frac{dt}{\left( (2t+1)^{2}+3 \right)^{2}}},$ now put $2t+1=\sqrt3u$ and the integral becomes $-\frac{8}{3\sqrt{3}}\int{\frac{du}{\left( u^{2}+1 \right)^{2}}}.$

As for solving the last integral, write the numerator as $(1+u^2)-u^2,$ and split the integral into two ones, integrate by parts the second one and we're done.

**simplependulum** · December 30th 2009, 08:13 PM

Here is another method ,

I write $4x^2 + 3$ in the form of

$A(2x+1) + B(x^2-1) +C(x^2+ x+1)$

I obtained $A= -7/3 , B=-2/3 ,C= 14/3$

the integral

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

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

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

sub. $x + 1/x = t$

you will be able to continue then ,

the last integral , you just need to make a trigo. substitution . ..

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