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Question Number 50384 by prof Abdo imad last updated on 16/Dec/18

f i n d \int \frac{d x}{(1 - x^{2}) (1 - x^{3})}

2) c a l c u l a t e \int_{2}^{\sqrt{5}} \frac{d x}{(1 - x^{2}) (1 - x^{3})}

Commented by Abdo msup. last updated on 23/Dec/18

l e t d e c o m p o s e F (x) = \frac{1}{(1 - x^{2}) (1 - x^{3})}

F (x) = \frac{1}{{(x - 1)}^{2} (x + 1) (x^{2} + x + 1)}

= \frac{a}{x - 1} + \frac{b}{{(x - 1)}^{2}} + \frac{c}{x + 1} + \frac{d x + f}{x^{2} + x + 1}

b = {l i m}_{x \to 1} {(x - 1)}^{2} F (x) = \frac{1}{6}

c = {l i m}_{x \to - 1} (x + 1) F (x) = \frac{1}{4} \Rightarrow

F (x) = \frac{a}{x - 1} + \frac{1}{6 {(x - 1)}^{2}} + \frac{1}{4 (x + 1)} + \frac{d x + f}{x^{2} + x + 1}

{l i m}_{x \to + \infty} x F (x) = a + \frac{1}{4} + d = 0 \Rightarrow a + d = - \frac{1}{4} \Rightarrow

F (x) = \frac{a}{x - 1} + \frac{1}{6 {(x - 1)}^{2}} + \frac{1}{4 (x + 1)} + \frac{(- a - \frac{1}{4}) x + f}{x^{2} + x + 1}

F (0) = 1 = - a + \frac{1}{6} + \frac{1}{4} + f \Rightarrow - a + f = 1 - \frac{1}{6} - \frac{1}{4}

= \frac{5}{6} - \frac{1}{4} = \frac{14}{24} = \frac{7}{12}

F (2) = a + \frac{1}{6} + \frac{1}{12} + \frac{(- 2 a - \frac{1}{2}) + f}{7} = \frac{1}{21} \Rightarrow

(1 - \frac{2}{7}) a + \frac{1}{4} - \frac{1}{14} + \frac{f}{7} = \frac{1}{21} \Rightarrow

\frac{5}{7} a + \frac{10}{56} + \frac{f}{7} = \frac{1}{21} \Rightarrow 5 a + \frac{70}{56} + f = \frac{1}{3} \Rightarrow

5 a + \frac{35}{28} + f = \frac{1}{3} \Rightarrow 5 a + f = \frac{1}{3} - \frac{35}{28}

b e c o n t i n u e d \dots . .

Answered by ajfour last updated on 16/Dec/18

I = \int \frac{d x}{(1 - x^{2}) (1 - x^{3})}

L e t \frac{1}{{(1 - x)}^{2} (1 + x) (1 + x + x^{2})}

= \frac{A}{{(1 - x)}^{2}} + \frac{B}{1 - x} + \frac{C}{1 + x} + \frac{D x + E}{1 + x + x^{2}}

A = \frac{1}{6}, C = \frac{1}{4}

1 = \frac{1}{6} (1 + x) (1 + x + x^{2})

+ B (1 + x) (1 - x^{3})

+ \frac{1}{4} (1 - x) (1 - x^{3})

+ (D x + E) {(1 - x)}^{2} (1 + x)

c o e f f . o f x^{4} = - B + \frac{1}{4} + D = 0

c o e f f . o f x^{3} = \frac{1}{6} - B - \frac{1}{4} + E - D = 0

c o n s t a n t t e r m c o e f f .

= \frac{1}{6} + B + \frac{1}{4} + E = 1

\dots .

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