partial fractions

**turtle** · January 21st 2007, 06:58 AM

I'm getting stuck on these, so if someone can please help, I'd really appreciate it.

Use partial fractions to evaluate the integral.

1. ∫(3xsquared - 2x + 12) / (xsquared + 4) squared

2. ∫ (2r) / (rsquared - 2r + 2)

3.∫ (theta) / (theta + 1)

Thanks!

**ThePerfectHacker** · January 21st 2007, 07:26 AM

Originally Posted by turtle

I'm getting stuck on these, so if someone can please help, I'd really appreciate it.

Use partial fractions to evaluate the integral.

1. ∫(3xsquared - 2x + 12) / (xsquared + 4) squared

$\frac{3x^2-2x+12}{(x^2+4)^2}=\frac{Ax+B}{x^2+4}+\frac{Cx+D}{( x^2+4)^2}$
Multiply through by the denominator,
$3x^2-2x+12=(Ax+B)(x^2+4)+(Cx+D)$
$3x^2-2x+12=Ax^3+Bx^2+4Ax+4B+Cx+D$
$3x^2-2x+12=Ax^3+Bx^2+(4A+C)x+(4B+D)$
Our eyes speaketh to us and say,
$A=0$
$B=3$
Thus,
$4A+C=4(0)+C=-2 \to C=-2$
$4B+D=4(3)+D=12 \to D=0$
Thus,
$\frac{3}{x^2+4}-\frac{2x}{(x^2+2)^2}$
The first integral,
$\int \frac{3}{x^2+4} dx = \frac{3}{4} \int \frac{1}{\frac{x^2}{4}+1} dx=\frac{3}{4} \int \frac{1}{1+(x/2)^2} dx = \frac{3}{4} \cdot 2 \cdot \tan^{-1} \frac{x}{2}+C$
The second integral,
$\int \frac{2x}{(x^2+2)^2}dx$
Let $u=x^2+2$
Thus,
$\int \frac{1}{u^2} du$
The rest is trivial.

**ThePerfectHacker** · January 21st 2007, 07:32 AM

Originally Posted by turtle

2. ∫ (2r) / (rsquared - 2r + 2)

You have,
$\int \frac{2x}{x^2-2x+2} dx$
Add an subtract,
$\int \frac{2x-2+2}{x^2-2x+2} dx$
Thus,
$\int \frac{2x-2}{x^2-2x+2} dx + \int \frac{2}{x^2-2x+2} dx$
The first integral is trivial, use $u=x^2-2x+2$ .
Thus, we get (details omitted),
$\ln |x^2-2x+2|+C$ .

The second integral needs some work, first we have,
$\int \frac{2}{x^2-2x+1+1} dx$
Complete the square,
$\int \frac{2}{(x-1)^2+1} dx$
The integral is trivial, use $u=x-1$ .
Thus,
$2\tan^{-1}(x-1)+C$
Thus,
$\ln |x^2-2x+2|+2\tan^{-1}(x-1)+C$
Note you can write,
$\ln (x^2-2x+2)+2\tan^{-1}(x-1)+C$
Becuase,
$x^2-2x+2>0$

**ThePerfectHacker** · January 21st 2007, 07:35 AM

Originally Posted by turtle

3.∫ (theta) / (theta + 1)

Add and subtract,
$\int \frac{x+1-1}{x+1} dx$
Thus,
$\int 1 dx - \int \frac{1}{x+1} dx$
The first integral is trivial,
$x+C$
The second integral is trivial, use $u=x+1$ .
Thus,
$x+\ln |x+1|+C$

**Soroban** · January 21st 2007, 08:18 AM

Hello, turtle!

Only the first one requires partial fractions . . .

$1)\;\int\frac{3x^2-2x+12}{(x^2+4)^2}\,dx$

We have: . $\frac{3x^2-2x+12}{(x^2+4)^2} \;=\;\frac{Ax + B}{x^2 + 4} + \frac{Cx+d}{(x^2+4)^2}$

Clear denominators: . $3x^2-2x+2\;=\;Ax(x^2+4) + B(x^2+4) + Cx + D$

$\begin{array}{cccc}\text{Let }x = 0\!: & 12\;=\;A\cdot0 + B\cdot4 + C\cdot0 + D \\\text{Let }x = 1\!: & 13\;=\;A\cdot5 + B\cdot5+ C + D \\ \text{Let }x = \text{-}1\!: & 17 \;=\;A(\text{-}5) + B\cdot5 - C + D \\ \text{Let }x = 2\!: & 20 \;=\;A\cdot16 + B\cdot8 + C\cdot2 + D \end{array}$

Solve the system of equations: . $\begin{array}{cccc} 4B + D \\ 5A + 5B + C + D \\ -5A + 5B - C + D \\ 16A + 8B + 2C + D\end{array} \begin{array}{cccc}=\\=\\=\\=\end{array} \begin{array}{cccc}12\\13\\17\\20\end{array}\; \begin{array}{cccc}(1)\\(2)\\(3)\\(4)\end{array}$

Add (2) and (3): . $10B + 2D\:=\:30\quad\Rightarrow\quad 5B + D \:=\:15$
. . . Subtract (1): . . . . . . . . . . . . . . . . . $4B + D \:=\:12$

We get: . $\boxed{B = 3}$ . . . hence: . $\boxed{D = 0}$

$\begin{array}{cc}\text{Substitute into (2):} & 5A + 5(3) + C + 0 \:=\:13 \\ \text{Substitute into (4):} & 16A + 8(3) + 2C + 0 \:=\:20 \end{array} \begin{array}{cc}\Rightarrow \\ \Rightarrow \end{array} \begin{array}{cc}5A + C \:=\:\text{-}2 \\ 8A + C\:=\:\text{-}2\end{array} \begin{array}{cc}(5)\\(6)\end{array} $

Subtract (5) from (6): . $3A \:=\:0\quad\Rightarrow\quad\boxed{ A = 0}$ . . . hence: . $\boxed{C = \text{-}2}$

The integral becomes: . $\int\left(\frac{3}{x^2+4} - \frac{2x}{(x^2+4)^2}\right)dx$

The first is of the $\arctan$ form.
For the second, let $u \:=\:x^2+4$

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