Integrating Rational Functions

[Mosaness]

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)

 

[Mark44]

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 [itex]expressions. It is causing what you wrote to not render correctly.<br /> <br /> <blockquote data-attributes="" data-quote="Mosaness" data-source="post: 4004137" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Mosaness said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> When we evaluate this, I get:<br /> <br /> $\frac{1}{2}$ln x - $\frac{1}{2}$tan^-1$\frac{x}{\sqrt{2}}$<br /> <br /> Or should it be:<br /> <br /> $\frac{1}{2}$ln x - $\frac{1}{2}$ln (x² + 2) </div> </div> </blockquote>[/itex]}

 

[Mosaness]

B was the coefficient of x in the numerator. What happened to x?

Also, don't mix ^{tags inside of $expressions. It is causing what you wrote to not render correctly.$}

^{$<br /> I fixed it:<br /> <br /> And I got <br /> <br /> \frac{1}{2}ln x - \frac{1}{4}ln(x² + 2)<br /> <br /> Is this correct?$}

 

[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.