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

Question Number 42085 by maxmathsup by imad last updated on 17/Aug/18

c a l c u l a t e \int_{1}^{+ \infty} \frac{2 x + 1}{3 + {(x + 1)}^{3}} d x

Commented by maxmathsup by imad last updated on 18/Aug/18

l e t A = \int_{1}^{+ \infty} \frac{2 x + 1}{3 + {(x + 1)}^{3}} d x c h a n g e m e n t x + 1 = t g i v e

A = \int_{2}^{+ \infty} \frac{2 (t - 1) + 1}{3 + t^{3}} d t = \int_{2}^{+ \infty} \frac{2 t - 1}{t^{3} + 3} d t a l s o c h a n g e m e n t^{3} \sqrt{3} u = t g i v e

A = \int_{\frac{2}{3_{\sqrt{3}}}}^{+ \infty} \frac{2 (^{3} \sqrt{3}) u - 1}{3 (1 + u^{3})}^{3} \sqrt{3} d u l e t α = \frac{2}{(^{3} \sqrt{3})}

= \frac{2}{3} {(^{3} \sqrt{3})}^{2} \int_{α}^{+ \infty} \frac{u}{u^{3} + 1} -^{3} \sqrt{3} \int_{α}^{+ \infty} \frac{d u}{u^{3} + 1}

= \frac{2}{3} {(\frac{2}{α})}^{2} \int_{α}^{+ \infty} \frac{u}{u^{3} + 1} d u - \frac{2}{α} \int_{α}^{+ \infty} \frac{d u}{u^{3} + 1}

= \int_{α}^{+ \infty} \frac{\frac{8}{3 α^{2}} u - \frac{2}{α}}{u^{3} + 1} d u = \frac{2}{α} \int_{α}^{+ \infty} \frac{\frac{4}{3 α} u - 1}{u^{3} + 1} d u

= \frac{2}{3 α^{2}} \int_{α}^{+ \infty} \frac{4 u - 3 α}{u^{3} + 1} d u l e t d e c o m p o s e F (u) = \frac{4 u - 3 α}{u^{3} + 1}

F (u) = \frac{4 u - 3 α}{(u + 1) (u^{2} - u + 1)} = \frac{a}{u + 1} + \frac{b u + c}{u^{2} - u + 1}

a = {l i m}_{u \to - 1} (u + 1) F (u) = \frac{- 4 - 3 α}{3}

{l i m}_{u \to + \infty} u F (u) = 0 = a + b \Rightarrow b = \frac{4 + 3 α}{3} \Rightarrow

F (u) = - \frac{3 α + 4}{3 (u + 1)} + \frac{1}{3} \frac{(3 α + 4) u + 3 c}{u^{2} - u + 1}

F (0) = - 3 α = - \frac{3 α + 4}{3} + c \Rightarrow c = - 3 α + \frac{3 α + 4}{3} = \frac{- 6 α + 4}{3} \Rightarrow

F (u) = - \frac{3 α + 4}{3 (u + 1)} + \frac{1}{3} \frac{(3 α + 4) u - 6 α + 4}{u^{2} - u + 1} \Rightarrow

\int_{α}^{+ \infty} F (u) d u = (- α - \frac{4}{3}) \int_{α}^{+ \infty} \frac{d u}{u + 1} + \frac{3 α + 4}{6} \int_{α}^{+ \infty} \frac{2 α - 1 + 1}{u^{2} - u + 1} d u

+ \frac{4 - 6 α}{3} \int_{α}^{+ \infty} \frac{d u}{u^{2} - u + 1}

= {[(- α - \frac{4}{3}) l n ∣ u + 1 ∣ + \frac{3 α + 4}{6} l n (u^{2} - u + 1)]}_{0}^{+ \infty} + \frac{4 - 6 α}{3} \int_{α}^{+ \infty} \frac{d u}{u^{2} - u + 1} \dots

Answered by tanmay.chaudhury50@gmail.com last updated on 18/Aug/18

t = x + 1 d t = d x

\int_{2}^{\infty} \frac{2 (t - 1) + 1}{3 + t^{3}} d t

\int_{2}^{\infty} \frac{2 t + 1}{3 + t^{3}} d t

\frac{2 t + 1}{t^{3} + {(3^{\frac{1}{3}})}^{3}} = \frac{2 t + 1}{(t + 3^{\frac{1}{3}}) (t^{2} - 3^{\frac{1}{3}} t + 3^{\frac{2}{3}})}

l e t a = 3^{\frac{1}{3}}

\frac{2 t + 1}{(t + a) (t^{2} - t a + a^{2})} = \frac{P}{t + a} + \frac{Q t + R}{(t^{2} - t a + a^{2})}

2 t + 1 = P (t^{2} - t a + a^{2}) + (t + a) (Q t + R)

2 t + 1 = t^{2} (P) + t (- P a) + ({P a}^{2}) + t^{2} (Q) + t (R) + t (Q a) + a R

2 t + 1 = t^{2} (P + Q) + t (- P a + R + Q a) + ({P a}^{2} + a R)

P + Q = 0

- P a + Q a + R = 2

{P a}^{2} + a R = 1

Q = - P

- P a - P a + \frac{1 - {P a}^{2}}{a} = 2

- 2 {P a}^{2} + 1 - {P a}^{2} = 2 a

- 3 {P a}^{2} = 2 a - 1

P = \frac{1 - 2 a}{3 a^{2}} Q = \frac{2 a - 1}{3 a^{2}}

R = \frac{1 - {P a}^{2}}{a} = \frac{1 - \frac{1 - 2 a}{3}}{a} = \frac{3 - 1 + 2 a}{3 a} = \frac{2 + 2 a}{3 a}

\int_{2}^{\infty} \frac{P}{t + a} + \frac{Q t + R}{t^{2} - t a + a^{2}} d t

P \int_{2}^{\infty} \frac{d t}{t + a} + \frac{Q}{2} \int_{2}^{\infty} \frac{2 t - a + a}{t^{2} - t a + a^{2}} + R \int_{2}^{\infty} \frac{d t}{t^{2} - 2 . t . \frac{a}{2} + \frac{a^{2}}{4} + \frac{3 a^{2}}{4}}

P \int_{2}^{\infty} \frac{d t}{t + a} + \frac{Q}{2} \int_{2}^{\infty} \frac{2 t - a}{t^{2} - t a + a^{2}} d t + (\frac{Q a}{2} + R) \int_{2}^{\infty} \frac{d t}{{(t - \frac{a}{2})}^{2} + {(\frac{a \sqrt{3}}{2})}^{2}}

P l n ∣ (t + a) ∣_{2}^{\infty} + \frac{Q}{2} l n ∣ (t^{2} - t a + a^{2}) ∣_{2}^{\infty} + (\frac{Q a}{2} + R) \times \frac{2}{a \sqrt{3}} ∣ {t a n}^{- 1} (\frac{t - \frac{a}{2}}{\frac{a \sqrt{3}}{2}}) ∣_{2}^{\infty}

t o b e c l m p l e t e d \dots