BeMath-x-dx-x-8-1-

Question Number 108073 by bemath last updated on 14/Aug/20

\frac{≜ B e M a t h ≜}{≺}

\int \frac{x d x}{x^{8} - 1} ?

Answered by john santu last updated on 14/Aug/20

\frac{JS ⊡}{≢}

I = \int \frac{x d x}{x^{8} - 1}

s e t x^{2} = z \Rightarrow 2 x d x = d z

I = \frac{1}{2} \int \frac{d z}{z^{4} - 1} = \frac{1}{2} \int \frac{d z}{(z^{2} - 1) (z^{2} + 1)}

I = \frac{1}{4} \int [\frac{1}{z^{2} - 1} - \frac{1}{z^{2} + 1}] d z

I = \frac{1}{8} \int (\frac{1}{z - 1} - \frac{1}{z + 1}) d z - \frac{1}{4} \tan^{- 1} (z) + c

I = \frac{1}{8} \ln ∣ \frac{z - 1}{z + 1} ∣ - \frac{1}{4} \tan^{- 1} (z) + c

I = \frac{1}{8} \ln ∣ \frac{x^{2} - 1}{x^{2} + 1} ∣ - \frac{1}{4} \tan^{- 1} (x^{2}) + c

Commented by bemath last updated on 14/Aug/20

w a w \dots c o o l l

Answered by malwaan last updated on 14/Aug/20

\int \frac{x}{x^{8} - 1} d x = \frac{1}{2} \int \frac{d (x^{2})}{{(x^{2})}^{4} - 1}

y = x^{2}

\Rightarrow \frac{1}{2} \int \frac{d y}{y^{4} - 1}

= \frac{1}{2} \int [\frac{d y}{(y^{2} + 1) (y + 1) (y - 1)}] d y

= \frac{1}{2} \int [\frac{a y + b}{y^{2} + 1} + \frac{c}{y + 1} + \frac{d}{y - 1}] d y

y = 1 \Rightarrow 4 d = 1 \Rightarrow d = \frac{1}{4}

y = - 1 \Rightarrow - 4 c = 1 \Rightarrow c = - \frac{1}{4}

y = 0 \Rightarrow - b - c + d = 1

\Rightarrow b = \frac{1}{2} - 1 = - 1

y = 2 \Rightarrow (2 a - \frac{1}{2}) (3) (1) + (- \frac{1}{4}) (5) (1)

+ (\frac{1}{4}) (5) (3) = 1 \Rightarrow 6 a = 0

\Rightarrow a = 0

∴ \frac{1}{2} \int [- \frac{1}{2} \frac{1}{y^{2} + 1} - \frac{1}{4} \frac{1}{y + 1} + \frac{1}{4} \frac{1}{y - 1}] d y

= - \frac{1}{4} {t a n}^{- 1} x^{2} - \frac{1}{8} l n ∣ x^{2} + 1 ∣

+ \frac{1}{8} l n ∣ x^{2} - 1 ∣ + C

Answered by mathmax by abdo last updated on 14/Aug/20

complex method z^{8} - 1 = 0 \Rightarrow z^{8} = e^{i 2 k π} \Rightarrow z_{k} = e^{\frac{i 2 k π}{8}} = e^{\frac{ik π}{4}}

k \in [[0, 7]] \Rightarrow F (z) = \frac{z}{z^{8} - 1} = \frac{z}{\prod_{k = 0}^{7} (z - z_{k})} = \sum_{k = 0}^{7} \frac{a_{k}}{z - z_{k}}

a_{k} = \frac{z_{k}}{{8 z}_{k}^{7}} = \frac{z_{k}^{2}}{8} \Rightarrow F (z) = \frac{1}{8} \sum_{k = 0}^{7} \frac{z_{k}^{2}}{z - z_{k}} \Rightarrow

\int F (z) dz = \frac{1}{8} \sum_{k = 0}^{7} z_{k}^{2} \ln (z - z_{k}) + C