Integrating Rational Functions

[Mosaness]

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)

 

[Mark44]

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)

 

[Mosaness]

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.}

<sup>I fixed it:

And I got

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

Is this correct?</sup>

 

[Mark44]

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.