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Integral

This document introduces an integral problem involving rational functions. It solves for the constants A, B, and C using boundary conditions when x=0, x=1, and x=2. The final solution is an integral of the form: 1 − ln(x2 + 3) + C 2

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andres David
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19 views3 pages

Integral

This document introduces an integral problem involving rational functions. It solves for the constants A, B, and C using boundary conditions when x=0, x=1, and x=2. The final solution is an integral of the form: 1 − ln(x2 + 3) + C 2

Uploaded by

andres David
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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1 Introduction

x2 − x + 6
Z
dx
x(x2 + 3)
Z Z
A Bx + C
dx + dx
x x2 + 3
 2 
x −x+6 A Bx + C
= + 2 x(x2 + 3)
x(x2 + 3) x x +3
x2 − x + 6 = A(x2 + 3) + (Bx + C)(x)

(1).
x=0
x − x + 6 = A(x2 + 3) + (Bx + C)(x)
2

(0)2 − 0 + 6 = A(02 + 3) + (Bx + C)0


6 = A(3) + 0 + 0
6 = 3A
6
=A
3
2=A

(2.)
x=1
x2 − x + 6 = A(x2 + 3) + (Bx + C)(x)
12 − 1 + 6 = 2(12 + 3) + (B1 + C)(1)
1 − 1 + 6 = 2(4) + B + c
6=8+B+C
6−8=B+C
−2 = B + C
−2 − C = B
(3.)
(X = 2)
2 − 2 + 6 = 2(22 + 3) + (B(2) + C)2
2

4 − 2 + 6 = 2(4 + 3) + (2B + C)2


8 = 2(7) + (2B + C)2

1
8 = 14 + 4B + 2C
−14 + 8 = 4B + 2C
−6 = 4B + 2C
Remplazamos B
−6 = 4(−2 − C) + 2C
−6 = −8 − 4C + 2C
−6 = −8 − 2C
−6 + 8 = −2C
2 = −2C
2
=C
−2
−1 = C
Remplazamos en la 2 ecuacion:

−2 − C = B

−2 − (−1) = B
−1 = B
2
x −x+6 A Bx + C
2
dx = dx + 2 dx
x(x + 3) x x +3
x2 − x + 6 A Bx c
dx = dx + 2 + 2 dx
x(x2 + 3) x x +3 x +3
x2 − x + 6 2 −1x −1
dx = dx + 2 dx + 2 dx
x(x2 + 3) x x +3 x +3
Z 2
x −x+6
Z Z Z
1 x 1
dx = 2 dx − dx − dx
x(x2 + 3) x x2 + 3 x2 + 3
 
1 2 1 x
2 ln |x| − ln(x + 3) − √ arctan √ +C
2 3 3
Z
x
− 2
dx
x +3

u = x2 + 3
du = 2x dx
du
= xdx
2
Z
du
−
2u

2
Z
1 1
− du
2 u
1
− ln(x2 + 3) + C
2
(1).
x=0
x2 − x + 6 = A(x2 + 3) + (Bx + C)(x)
(0)2 − 0 + 6 = A(02 + 3) + (Bx + C)0
6 = A(3) + 0 + 0
6 = 3A
6
=A
3
2=A
x2 − x + 6
dx
x(x2 + 3)

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