dx-x-3-x-2-x-1-

Question Number 121300 by liberty last updated on 06/Nov/20

\int \frac{dx}{x^{3} + x^{2} + x + 1} ?

Commented by liberty last updated on 06/Nov/20

Answered by TANMAY PANACEA last updated on 06/Nov/20

\int \frac{d x}{x^{2} (x + 1) + 1 (x + 1)}

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

1 = a (x^{2} + 1) + (x + 1) (b x + c)

1 = {a x}^{2} + a + {b x}^{2} + c x + b x + c

1 = x^{2} (a + b) + x (b + c) + (a + c)

a + b = 0

b + c = 0

a + c = 1

a + c = 1

b = - a

c - a = 0 \to a = c

a = \frac{1}{2} b = \frac{- 1}{2} c = \frac{1}{2}

\frac{1}{2} \int \frac{d x}{x + 1} + \frac{1}{2} \int \frac{- x + 1}{x^{2} + 1} d x

\frac{1}{2} \int \frac{d x}{x + 1} - \frac{1}{4} \int \frac{2 x - 2}{x^{2} + 1} d x

\frac{1}{2} \int \frac{d x}{x + 1} - \frac{1}{4} \int \frac{d (x^{2} + 1)}{x^{2} + 1} + \frac{1}{2} \int \frac{d x}{x^{2} + 1}

\frac{1}{2} l n (x + 1) - \frac{1}{4} l n (x^{2} + 1) + \frac{1}{2} {t a n}^{- 1} (x) + c

Answered by Bird last updated on 06/Nov/20

I = \int \frac{d x}{1 + x + x^{2} + x^{3}} \Rightarrow

I = \int \frac{d x}{\frac{1 - x^{4}}{1 - x}} = \int \frac{1 - x}{1 - x^{4}} d x

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

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

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

F (x) = \frac{a}{x + 1} + \frac{b x + c}{x^{2} + 1}

a = \frac{1}{2}, {l i m}_{x \to + \infty} x F (x) = 0 = a + b \Rightarrow

b = - \frac{1}{2}

F (o) = 1 = a + c \Rightarrow c = 1 - \frac{1}{2} = \frac{1}{2} \Rightarrow

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

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

I = \frac{1}{2} \int \frac{d x}{x + 1} - \frac{1}{4} \int \frac{2 x}{x^{2} + 1} d x + \frac{1}{2} \int \frac{d x}{x^{2} + 1}

= \frac{1}{2} l n ∣ x + 1 ∣ - \frac{1}{4} l n (x^{2} + 1) + \frac{1}{2} a r c t a n x + C

Facebook Tweet Pin