# Integrating Rational Functions

1. ∫$\frac{dx}{x3 + 2x}$

We're suppose to evaluate the integral.

Use Partial Fraction Decomposition:

$\frac{1}{x3 + 2x}$ = $\frac{A}{x}$ + $\frac{Bx + C}{x2 + 2}$

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

Solving for A gives $\frac{1}{2}$
Solving for B gives -$\frac{1}{2}$
Solving for C gives 0

∫$\frac{dx}{x(x2 + 2}$ = $\frac{1}{2}$∫$\frac{dx}{x}$ - $\frac{1}{2}$∫$\frac{dx}{x2 + 2}$

When we evaluate this, I get:

$\frac{1}{2}$ln x - $\frac{1}{2}$tan-1$\frac{x}{\sqrt{2}}$

Or should it be:

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

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Mark44
Mentor
1. ∫$\frac{dx}{x3 + 2x}$

We're suppose to evaluate the integral.

Use Partial Fraction Decomposition:

$\frac{1}{x3 + 2x}$ = $\frac{A}{x}$ + $\frac{Bx + C}{x2 + 2}$

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

Solving for A gives $\frac{1}{2}$
Solving for B gives -$\frac{1}{2}$
Solving for C gives 0

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

Also, don't mix tags inside of $expressions. It is causing what you wrote to not render correctly. When we evaluate this, I get: [itex]\frac{1}{2}$ln x - $\frac{1}{2}$tan-1$\frac{x}{\sqrt{2}}$

Or should it be:

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

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?

Mark44
Mentor
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.