∫_3 ^6 ((x+1)/(x^3 +x^2 −6x))dx


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Question Number 113867 by Ar Brandon last updated on 15/Sep/20

$\int_{3}^{6} \frac{x + 1}{x^{3} + x^{2} - 6 x} dx$
	Answered by abdomsup last updated on 15/Sep/20
	$I =∫_3 ^6 ((x+1)/(x^3 +x^2 −6x))dx let decompise F(x)=((x+1)/(x^3 +x^2 −6x)) ⇒ F(x) =((x+1)/(x(x^2 +x−6))) x^2 +x−6=0→Δ=1+24=25 x_1 =((−1+5)/2)=2 and x_2 =((−1−5)/2)=−3 ⇒F(x)=((x+1)/(x(x−2)(x+3))) =(a/x) +(b/(x−2)) +(c/(x+3)) a=−(1/6) , b=(3/(2(6))) =(1/4) c=((−2)/((−3)(−5))) =(2/(15)) ⇒ F(x)=−(1/(6x)) +(1/(4(x−2))) +(2/(15(x+3))) ⇒I =∫_3 ^6 (−(1/(6x)) +(1/(4(x−2)))+(2/(15(x+3))))dx =−(1/6)[ln∣x∣]_3 ^6 +(1/4)[ln∣x−2∣]_3 ^6 +(2/(15))[ln∣x+3∣]_3 ^6 =−(1/6)ln(2)+(1/4)(2ln2) +(2/(15)){2ln(3)−ln(6)} =((1/2)−(1/6))ln(2)+(2/(15)){ln(3)−ln(2)} =((1/3)−(2/(15)))ln(2)+(2/(15))ln(3) =(1/5)ln(2)+(2/(15))ln(3)$
	$I = \int_{3}^{6} \frac{x + 1}{x^{3} + x^{2} - 6 x} d x l e t d e c o m p i s e$ $F (x) = \frac{x + 1}{x^{3} + x^{2} - 6 x} \Rightarrow$ $F (x) = \frac{x + 1}{x (x^{2} + x - 6)}$ $x^{2} + x - 6 = 0 \to Δ = 1 + 24 = 25$ $x_{1} = \frac{- 1 + 5}{2} = 2 a n d x_{2} = \frac{- 1 - 5}{2} = - 3$ $\Rightarrow F (x) = \frac{x + 1}{x (x - 2) (x + 3)}$ $= \frac{a}{x} + \frac{b}{x - 2} + \frac{c}{x + 3}$ $a = - \frac{1}{6}, b = \frac{3}{2 (6)} = \frac{1}{4}$ $c = \frac{- 2}{(- 3) (- 5)} = \frac{2}{15} \Rightarrow$ $F (x) = - \frac{1}{6 x} + \frac{1}{4 (x - 2)} + \frac{2}{15 (x + 3)}$ $\Rightarrow I = \int_{3}^{6} (- \frac{1}{6 x} + \frac{1}{4 (x - 2)} + \frac{2}{15 (x + 3)}) d x$ $= - \frac{1}{6} {[l n ∣ x ∣]}_{3}^{6} + \frac{1}{4} {[l n ∣ x - 2 ∣]}_{3}^{6}$ $+ \frac{2}{15} {[l n ∣ x + 3 ∣]}_{3}^{6}$ $= - \frac{1}{6} l n (2) + \frac{1}{4} (2 l n 2)$ $+ \frac{2}{15} {2 l n (3) - l n (6)}$ $= (\frac{1}{2} - \frac{1}{6}) l n (2) + \frac{2}{15} {l n (3) - l n (2)}$ $= (\frac{1}{3} - \frac{2}{15}) l n (2) + \frac{2}{15} l n (3)$ $= \frac{1}{5} l n (2) + \frac{2}{15} l n (3)$
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