find ∫ ((3x+2)/((x+1)^2 (x−2)^3 ))dx


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 73479 by abdomathmax last updated on 13/Nov/19

$f i n d \int \frac{3 x + 2}{{(x + 1)}^{2} {(x - 2)}^{3}} d x$
Commented by abdomathmax last updated on 17/Nov/19
$let decompose the fraction F(x)=((3x+2)/((x+1)^2 (x−2)^3 )) F(x)=(a/(x+1)) +(b/((x+1)^2 )) +(c/(x−2)) +(d/((x−2)^2 )) +(e/((x−2)^3 )) b =(x+1)^2 F(x)∣_(x=−1) =((−1)/(−27)) =(1/(27)) e=(x−2)^3 F(x)∣_(x=2) = (8/(27)) ⇒ F(x)=(a/(x+1))+(1/(27(x+1)^2 )) +(c/(x−2)) +(d/((x−2)^2 )) +(8/(27(x−2)^3 )) lim_(x→+∞) xF(x)=0 =a+c ⇒c=−a ⇒ F(x) =(a/(x+1)) +(1/(27(x+1)^2 ))−(a/(x−2)) +(d/((x−2)^2 )) +(8/(27(x−2)^3 )) F(0) =−(1/4) =a +(1/(27)) +(a/2) +(d/4) −(1/(27)) ⇒ −(1/4) =(3/2)a +(d/4) ⇒−1=6a+d F(1) =(5/(4(−1))) =−(5/4) =(a/2) +(1/(4×27)) +a+d−(8/(27)) =(3/2)a +d +((1/4)−8)(1/(27)) ⇒ −5 =6a +4d+ (1−32)×(1/(27)) =6a+4d −((31)/(27)) 6a+4d =((31)/(27))−5 =((31−135)/(27)) =((104)/(27)) ⇒ 3a+2d =((52)/(27)) we get the system { ((6a+d=−1)),((3a+2d=((52)/(27)))) :} Δ_s =12−3 =9 a=( determinant (((−1 1)),((((52)/(27 )) 2)))/9) =((−2−((52)/(27)))/9) =((−54−52)/(27×9)) =... b=( determinant (((6 −1)),((3 ((52)/(27)))))/9) =((((6×52)/(27))+3)/9) =... the coefficients are known ⇒ ∫ F(x)dx =aln∣x+1∣−(1/(27(x+1))) −aln∣x−2∣−(d/(x−2)) −(4/(27))(1/((x−2)^2 )) +C.$
$l e t d e c o m p o s e t h e f r a c t i o n F (x) = \frac{3 x + 2}{{(x + 1)}^{2} {(x - 2)}^{3}}$ $F (x) = \frac{a}{x + 1} + \frac{b}{{(x + 1)}^{2}} + \frac{c}{x - 2} + \frac{d}{{(x - 2)}^{2}} + \frac{e}{{(x - 2)}^{3}}$ $b = {(x + 1)}^{2} F (x) ∣_{x = - 1} = \frac{- 1}{- 27} = \frac{1}{27}$ $e = {(x - 2)}^{3} F (x) ∣_{x = 2} = \frac{8}{27} \Rightarrow$ $F (x) = \frac{a}{x + 1} + \frac{1}{27 {(x + 1)}^{2}} + \frac{c}{x - 2} + \frac{d}{{(x - 2)}^{2}} + \frac{8}{27 {(x - 2)}^{3}}$ ${l i m}_{x \to + \infty} x F (x) = 0 = a + c \Rightarrow c = - a \Rightarrow$ $F (x) = \frac{a}{x + 1} + \frac{1}{27 {(x + 1)}^{2}} - \frac{a}{x - 2} + \frac{d}{{(x - 2)}^{2}} + \frac{8}{27 {(x - 2)}^{3}}$ $F (0) = - \frac{1}{4} = a + \frac{1}{27} + \frac{a}{2} + \frac{d}{4} - \frac{1}{27} \Rightarrow$ $- \frac{1}{4} = \frac{3}{2} a + \frac{d}{4} \Rightarrow - 1 = 6 a + d$ $F (1) = \frac{5}{4 (- 1)} = - \frac{5}{4} = \frac{a}{2} + \frac{1}{4 \times 27} + a + d - \frac{8}{27}$ $= \frac{3}{2} a + d + (\frac{1}{4} - 8) \frac{1}{27} \Rightarrow$ $- 5 = 6 a + 4 d + (1 - 32) \times \frac{1}{27} = 6 a + 4 d - \frac{31}{27}$ $6 a + 4 d = \frac{31}{27} - 5 = \frac{31 - 135}{27} = \frac{104}{27} \Rightarrow$ $3 a + 2 d = \frac{52}{27} w e g e t t h e s y s t e m {\begin{cases} 6 a + d = - 1 \\ 3 a + 2 d = \frac{52}{27} \end{cases}$ $Δ_{s} = 12 - 3 = 9$ $a = \frac{\| \begin{matrix} - 1 1 \\ \frac{52}{27} 2 \end{matrix} \|}{9} = \frac{- 2 - \frac{52}{27}}{9} = \frac{- 54 - 52}{27 \times 9} = . . .$ $b = \frac{\| \begin{matrix} 6 - 1 \\ 3 \frac{52}{27} \end{matrix} \|}{9} = \frac{\frac{6 \times 52}{27} + 3}{9} = . . .$ $t h e c o e f f i c i e n t s a r e k n o w n \Rightarrow$ $\int F (x) d x = a l n ∣ x + 1 ∣ - \frac{1}{27 (x + 1)} - a l n ∣ x - 2 ∣ - \frac{d}{x - 2}$ $- \frac{4}{27} \frac{1}{{(x - 2)}^{2}} + C .$
	Answered by MJS last updated on 13/Nov/19

	$\int \frac{3 x + 2}{{(x - 2)}^{3} {(x + 1)}^{2}} d x =$ $= \frac{8}{9} \int \frac{d x}{{(x - 2)}^{3}} - \frac{7}{27} \int \frac{d x}{{(x - 2)}^{2}} + \frac{2}{27} \int \frac{d x}{x - 2} + \frac{1}{27} \int \frac{d x}{{(x + 1)}^{2}} - \frac{2}{27} \int \frac{d x}{x + 1} =$ $= - \frac{4}{9 {(x - 2)}^{2}} + \frac{7}{27 (x - 2)} + \frac{2}{27} \ln (x - 2) - \frac{1}{27 (x + 1)} - \frac{2}{27} \ln (x + 1) =$ $= \frac{2 x^{2} - 5 x - 10}{9 {(x - 2)}^{2} (x + 1)} + \frac{2}{27} \ln ∣ \frac{x - 2}{x + 1} ∣ + C$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum