Integrating Rational Functions

  • Thread starter Mosaness
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  • #1
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1. ∫[itex]\frac{dx}{x3 + 2x}[/itex]

We're suppose to evaluate the integral.

Use Partial Fraction Decomposition:

[itex]\frac{1}{x3 + 2x}[/itex] = [itex]\frac{A}{x}[/itex] + [itex]\frac{Bx + C}{x2 + 2}[/itex]

1 = A(x2 + 2) + (Bx + C)(x)
1 = Ax2 + 2A + Bx2 + Cx
1 = x2( A + B) + Cx + 2A

Solving for A gives [itex]\frac{1}{2}[/itex]
Solving for B gives -[itex]\frac{1}{2}[/itex]
Solving for C gives 0

∫[itex]\frac{dx}{x(x2 + 2}[/itex] = [itex]\frac{1}{2}[/itex]∫[itex]\frac{dx}{x}[/itex] - [itex]\frac{1}{2}[/itex]∫[itex]\frac{dx}{x2 + 2}[/itex]

When we evaluate this, I get:

[itex]\frac{1}{2}[/itex]ln x - [itex]\frac{1}{2}[/itex]tan-1[itex]\frac{x}{\sqrt{2}}[/itex]

Or should it be:

[itex]\frac{1}{2}[/itex]ln x - [itex]\frac{1}{2}[/itex]ln (x2 + 2)
 

Answers and Replies

  • #2
33,722
5,418
1. ∫[itex]\frac{dx}{x3 + 2x}[/itex]

We're suppose to evaluate the integral.

Use Partial Fraction Decomposition:

[itex]\frac{1}{x3 + 2x}[/itex] = [itex]\frac{A}{x}[/itex] + [itex]\frac{Bx + C}{x2 + 2}[/itex]

1 = A(x2 + 2) + (Bx + C)(x)
1 = Ax2 + 2A + Bx2 + Cx
1 = x2( A + B) + Cx + 2A

Solving for A gives [itex]\frac{1}{2}[/itex]
Solving for B gives -[itex]\frac{1}{2}[/itex]
Solving for C gives 0

∫[itex]\frac{dx}{x(x2 + 2}[/itex] = [itex]\frac{1}{2}[/itex]∫[itex]\frac{dx}{x}[/itex] - [itex]\frac{1}{2}[/itex]∫[itex]\frac{dx}{x2 + 2}[/itex]
B was the coefficient of x in the numerator. What happened to x?

Also, don't mix tags inside of [itex] expressions. It is causing what you wrote to not render correctly.

When we evaluate this, I get:

[itex]\frac{1}{2}[/itex]ln x - [itex]\frac{1}{2}[/itex]tan-1[itex]\frac{x}{\sqrt{2}}[/itex]

Or should it be:

[itex]\frac{1}{2}[/itex]ln x - [itex]\frac{1}{2}[/itex]ln (x2 + 2)
 
  • #3
92
0
B was the coefficient of x in the numerator. What happened to x?

Also, don't mix tags inside of [itex] expressions. It is causing what you wrote to not render correctly.


I fixed it:

And I got

\frac{1}{2}ln x - \frac{1}{4}ln(x2 + 2)

Is this correct?
 
  • #4
33,722
5,418
Why don't you check for yourself? If your answer is correct, it should be true that
d/dx[1/2 * ln(x) - (1/4) * ln(x2 + 2)] = 1/(x3 + 2x), the original integrand.
 

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