fracciones parciales

Ejercicios propuestos en clase

$\int \frac{2 d x}{(x - 2) (x + 4)}$
$\int \frac{d x}{(x + 5) (x + 6)}$
$\int \frac{d x}{(x - 3) (x + 4)}$
$\int \frac{d x}{(x - 5) (x + 6)}$

Soluciones:

$\int \frac{2 d x}{(x - 2) (x + 4)}$ $\frac{2}{(x - 2) (x + 4)} = \frac{A}{x - 2} + \frac{B}{x + 4} = \frac{A (x + 4) + B (x - 2)}{(x - 2) (x + 4)}$ $⟹ \frac{A x + 4 A + B x - 2 B}{(x - 2) (x + 4)} = \frac{(A + B) x + (4 A - 2 B)}{(x - 2) (x + 4)}$ $∴ \frac{2}{(x - 2) (x + 4)} = \frac{(A + B) x + (4 A - 2 B)}{(x - 2) (x + 4)}$ $2 = (A + B) x + (4 A - 2 B)$

⟹ \begin{array}{r} A & + & B & = & 0 \\ 4 A & - & 2 B & = & 2 \end{array}

∴ 6 A = 2 ⟹ A = \frac{1}{3} \land B = - \frac{1}{3}

⟹ \frac{2}{(x - 2) (x + 4)} = \frac{A}{x - 2} + \frac{B}{x + 4}

\int \frac{2 d x}{(x - 2) (x + 4)} = \int [\frac{A}{x - 2} + \frac{B}{x + 4}] d x

⟹ \int \frac{A}{x - 2} d x + \int \frac{B}{x + 4} d x

A \int \frac{d x}{x - 2} + B \int \frac{d x}{x + 4}

(\frac{1}{3}) L n | x - 2 | + C_{1} + (- \frac{1}{3}) L n | x + 4 | + C_{2}

(\frac{1}{3}) L n | x - 2 | - (\frac{1}{3}) L n | x + 4 | + C

(\frac{1}{3}) (L n | x - 2 | - L n | x + 4 |) + C

\frac{1}{3} (L n \frac{| x - 2 |}{| x + 4 |}) + C

$\int \frac{d x}{(x + 5) (x + 6)}$

\frac{1}{(x + 5) (x + 6)} = \frac{A}{x + 5} + \frac{B}{x + 6} = \frac{A (x + 6) + B (x + 5)}{(x + 5) (x + 6)}

⟹ \frac{A x + 6 A + B x + 5 B}{(x + 5) (x + 6)} = \frac{(A + B) x + (6 A + 5 B)}{(x + 5) (x + 6)}

∴ \frac{1}{(x + 5) (x + 6)} = \frac{(A + B) x + (6 A + 5 B)}{(x + 5) (x + 6)}

1 = (A + B) x + (6 A + 5 B)

⟹ \begin{array}{r} A & + & B & = & 0 \\ 6 A & + & 5 B & = & 1 \end{array}

∴ A = 1 \land B = - 1

⟹ \frac{1}{(x + 5) (x + 6)} = \frac{A}{x + 5} + \frac{B}{x + 6}

\int \frac{1}{(x + 5) (x + 6)} = \int [\frac{A}{(x + 5)} + \frac{B}{(x + 6)}] d x

⟹ \int \frac{A}{x + 5} d x + \int \frac{B}{x + 6} d x

A \int \frac{d x}{x + 5} + B \int \frac{d x}{x + 6}

(1) L n | x + 5 | + C_{1} + (- 1) L n | x + 6 | + C_{2}

L n | x + 5 | - L n | x + 6 | + C

L n \frac{| x + 5 |}{| x + 6 |} + C

$\int \frac{d x}{(x - 3) (x + 4)}$

\frac{1}{(x - 3) (x + 4)} = \frac{A}{x - 3} + \frac{B}{x + 4} = \frac{A (x + 4) + B (x - 3)}{(x - 3) (x + 4)}

⟹ \frac{A x + 4 A + B x - 3 B}{(x - 3) (x + 4)} = \frac{(A + B) x + (4 A - 3 B)}{(x - 3) (x + 4)}

∴ \frac{1}{(x - 3) (x + 4)} = \frac{(A + B) x + (4 A - 3 B)}{(x - 3) (x + 4)}

1 = (A + B) x + (4 A - 3 B)

⟹ \begin{array}{r} A & + & B & = & 0 \\ 4 A & - & 3 B & = & 1 \end{array}

∴ A = \frac{1}{7} \land B = - \frac{1}{7}

⟹ \frac{1}{(x - 3) (x + 4)} = \frac{A}{x - 3} + \frac{B}{x + 4}

\int \frac{1}{(x - 3) (x + 4)} = \int [\frac{A}{(x - 3)} + \frac{B}{(x + 4)}] d x

⟹ \int \frac{A}{x - 3} d x + \int \frac{B}{x + 4} d x

A \int \frac{d x}{x - 3} + B \int \frac{d x}{x + 4}

(\frac{1}{7}) L n | x - 3 | + C_{1} + (- \frac{1}{7}) L n | x + 4 | + C_{2}

(\frac{1}{7}) (L n | x - 3 | - L n | x + 4 |) + C

\frac{1}{7} (L n \frac{| x - 3 |}{| x + 4 |}) + C

$\int \frac{d x}{(x - 5) (x + 6)}$

\frac{1}{(x - 5) (x + 6)} = \frac{A}{x - 5} + \frac{B}{x + 6} = \frac{A (x + 6) + B (x - 5)}{(x - 5) (x + 6)}

⟹ \frac{A x + 6 A + B x - 5 B}{(x - 5) (x + 6)} = \frac{(A + B) x + (6 A - 5 B)}{(x - 5) (x + 6)}

∴ \frac{1}{(x - 5) (x + 6)} = \frac{(A + B) x + (6 A - 5 B)}{(x - 5) (x + 6)}

1 = (A + B) x + (6 A - 5 B)

⟹ \begin{array}{r} A & + & B & = & 0 \\ 6 A & - & 5 B & = & 1 \end{array}

∴ A = \frac{1}{11} \land B = - \frac{1}{11}

⟹ \frac{1}{(x - 5) (x + 6)} = \frac{A}{x - 5} + \frac{B}{x + 6}

\int \frac{1}{(x - 5) (x + 6)} = \int [\frac{A}{(x - 5)} + \frac{B}{(x + 6)}] d x

⟹ \int \frac{A}{x - 5} d x + \int \frac{B}{x + 6} d x

A \int \frac{d x}{x - 5} + B \int \frac{d x}{x + 6}

(\frac{1}{11}) L n | x - 5 | + C_{1} + (- \frac{1}{11}) L n | x + 6 | + C_{2}

(\frac{1}{11}) (L n | x - 5 | - L n | x + 6 |) + C

\frac{1}{11} (L n \frac{| x - 5 |}{| x + 6 |}) + C

De Forma general, para:

\int \frac{A d x}{(B x + C) (D x + E)}

Tenemos:

\frac{A}{(B x + C) (D x + E)} = \frac{F}{(B x + C)} + \frac{H}{(D x + E)}

\frac{A}{(B x + C) (D x + E)} = \frac{F (D x + E) + H (B x + C)}{(B x + C) (D x + E)}

⟹ \frac{A}{(B x + C) (D x + E)} = \frac{(F D + H B) x + (F E + H C)}{(B x + C) (D x + E)}

A = (F D + H B) x + (F E + H C)

⟹ \begin{array}{r} F D & + & H B & = & 0 \\ F E & + & H C & = & A \end{array}