1-x-3-1-dx-

Question Number 49032 by AdqhK ÐQeQqQ last updated on 01/Dec/18

\int \frac{1}{x^{3} + 1} d x = ? ?

Answered by arvinddayama01@gmail.com. last updated on 01/Dec/18

\int \frac{1}{(x + 1) (x^{2} - x + 1)} d x ∵ a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})

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

1 = {A x}^{2} - A x + A + {B x}^{2} + B x + C x + C

A + B = 0

- A + B + C = 0

A + C = 1

A = 1 / 3 B = - 1 / 3 c = 2 / 3

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

= \frac{1}{3} l n (x + 1) - \frac{1}{6} \int \frac{2 x - 1 - 3}{x^{2} - x + 1} d x

= \frac{1}{3} l n (x + 1) - \frac{1}{6} \int \frac{2 x - 1}{x^{2} - x + 1} d x + \frac{3}{6} \int \frac{1}{{(x - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d x

= \frac{1}{3} l n (x + 1) - \frac{1}{6} l n (x^{2} - x + 1) + \frac{1}{2} . \frac{1}{\sqrt{3} / 2} {t a n}^{- 1} (\frac{x - \frac{1}{2}}{\frac{\sqrt{3}}{2}}) + C

= \frac{1}{3} l n (x + 1) - \frac{1}{3} l n (\sqrt{x^{2} - x + 1}) + \frac{1}{\sqrt{3}} {t a n}^{- 1} (\frac{2 x - 1}{\sqrt{3}}) + C

= \frac{1}{3} l n (\frac{x + 1}{\sqrt{x^{2} - x + 1}}) + \frac{1}{\sqrt{3}} {t a n}^{- 1} (\frac{2 x - 1}{\sqrt{3}}) + C

Commented by AdqhK ÐQeQqQ last updated on 02/Dec/18

t h a n k u v e r y m u c h s i r

Answered by hknkrc46 last updated on 01/Dec/18

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

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

\int \frac{1}{x^{3} + 1} d x = \int (\frac{A}{x + 1} + \frac{B x + C}{x^{2} - x + 1}) d x ★

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

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

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

{1 = A (x^{2} - x + 1) + (B x + C) (x + 1)}

{1 = {A x}^{2} - A x + A + {B x}^{2} + B x + C x + C}

{1 = (A + B) x^{2} + (B + C - A) x + C + A}

{A + B = 0, B + C - A = 0, C + A = 1}

{A = \frac{1}{3}, B = - \frac{1}{3}, C = \frac{2}{3}}

★ \int \frac{1}{x^{3} + 1} d x = \int (\frac{A}{x + 1} + \frac{B x + C}{x^{2} - x + 1}) d x

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

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

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

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

= \frac{1}{3} \int (\frac{1}{x + 1} - \frac{1}{2} \cdot \frac{2 x - 1}{x^{2} - x + 1} + \frac{3}{2} \cdot \frac{1}{{(x - \frac{1}{2})}^{2} + \frac{3}{4}}) d x

= \frac{1}{3} \int (\frac{1}{x + 1} - \frac{1}{2} \cdot \frac{2 x - 1}{x^{2} - x + 1} + \frac{3}{2} \cdot \frac{1}{{(x - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}) d x

= \frac{1}{3} \int \frac{d x}{x + 1} - \frac{1}{6} \int \frac{2 x - 1}{x^{2} - x + 1} d x + \frac{1}{2} \int \frac{d x}{{(x - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}

\int \frac{1}{x^{3} + 1} d x = I_{1} + I_{2} + I_{3} ★ ★

{I_{1} = \frac{1}{3} \int \frac{d x}{x + 1}}

{I_{2} = - \frac{1}{6} \int \frac{2 x - 1}{x^{2} - x + 1} d x}

{I_{3} = \frac{1}{2} \int \frac{d x}{{(x - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}}

I_{1} = \frac{1}{3} \int \frac{d x}{x + 1} = \frac{1}{3} \int \frac{\frac{d}{d x} (x + 1)}{x + 1} d x = \frac{1}{3} \ln ∣ x + 1 ∣ + c_{1}

I_{2} = - \frac{1}{6} \int \frac{2 x - 1}{x^{2} - x + 1} d x = - \frac{1}{6} \int \frac{\frac{d}{d x} (x^{2} - x + 1)}{x^{2} - x + 1} d x = - \frac{1}{6} \ln ∣ x^{2} - x + 1 ∣ + c_{2}

I_{3} = \frac{1}{2} \int \frac{d x}{{(x - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} \to {\begin{cases} x - \frac{1}{2} = \frac{\sqrt{3}}{2} \tan α \\ d x = \frac{\sqrt{3}}{2} \sec^{2} α d α \\ α = \tan^{- 1} (\frac{2 x - 1}{\sqrt{3}}) \end{cases}

I_{3} = \frac{1}{2} \int \frac{\frac{\sqrt{3}}{2} \sec^{2} α}{{(\frac{\sqrt{3}}{2})}^{2} \sec^{2} α} d α = \frac{1}{\sqrt{3}} \int d α = \frac{1}{\sqrt{3}} α + c_{3}

I_{3} = \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{2 x - 1}{\sqrt{3}}) + c_{3}

★ ★ \int \frac{1}{x^{3} + 1} d x = I_{1} + I_{2} + I_{3}

\int \frac{1}{x^{3} + 1} d x = \frac{1}{3} \ln ∣ x + 1 ∣ + c_{1} - \frac{1}{6} \ln ∣ x^{2} - x + 1 ∣ + c_{2} + \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{2 x - 1}{\sqrt{3}}) + c_{3}

\int \frac{1}{x^{3} + 1} d x = \frac{1}{3} \ln ∣ x + 1 ∣ - \frac{1}{6} \ln ∣ x^{2} - x + 1 ∣ + \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{2 x - 1}{\sqrt{3}}) + c_{1} + c_{2} + c_{3}

♣ c_{1} + c_{2} + c_{3} = c

⧫ a \neq 0 \land \forall x \in R \land △_{({a x}^{2} + b x + c)} = b^{2} - 4 a c < 0 \Rightarrow {\begin{cases} {a x}^{2} + b x + c > 0 \land ∣ {a x}^{2} + b x + c ∣= {a x}^{2} + b x + c, a > 0 \\ {a x}^{2} + b x + c < 0 \land ∣ {a x}^{2} + b x + c ∣= - ({a x}^{2} + b x + c), a < 0 \end{cases}

⧫ △_{x^{2} - x + 1} = {(- 1)}^{2} - 4 \cdot 1 \cdot 1 = - 3 < 0 \Rightarrow∣ x^{2} - x + 1 ∣= x^{2} - x + 1 (a = 1 > 0)

\int \frac{1}{x^{3} + 1} d x = \frac{1}{3} \ln ∣ x + 1 ∣ - \frac{1}{6} \ln (x^{2} - x + 1) + \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{2 x - 1}{\sqrt{3}}) + c

Facebook Tweet Pin