find ∫_2 ^3 ((2x^2 +3)/((x−1)^2 (x^2 +1))) dx


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Question Number 34911 by abdo imad last updated on 12/May/18

$f i n d \int_{2}^{3} \frac{2 x^{2} + 3}{{(x - 1)}^{2} (x^{2} + 1)} d x$
Commented by math khazana by abdo last updated on 13/May/18
$let?decompose F(x)= ((2x^2 +3)/((x−1)^2 (x^2 +1))) F(x) = (a/(x−1)) +(b/((x−1)^2 )) +((cx+d)/(x^2 +1)) b=lim_(x→1) (x−1)^2 F(x)= (5/2) lim_(x→+∞) xF(x)= 0= a +c ⇒c=−a ⇒ F(x)= (a/(x−1)) +(5/(2(x−1)^2 )) +((−ax +d)/(x^2 +1)) F(o) =3 = −a +(5/2) +d ⇒d =3 +a−(5/2) ⇒ F(x)= (a/(x−1)) +(5/(2(x−1)^2 )) +((−ax +a +(1/2))/(x^2 +1)) F(2) = a +(5/2) +((−a +(1/2))/5) = ((11)/5) ⇒ 5a +((25)/2) −a +(1/2) =11 ⇒4a +13=11 ⇒ 4a = −2 ⇒a =−(1/2) ⇒c=(1/2) and d =3−(1/2) −(5/2)=0 F(x) = ((−1)/(2(x−1))) +(5/(2(x−1)^2 )) + (((1/2)x)/(x^2 +1)) ∫ F(x)dx = −(1/2)ln∣x−1∣ −(5/(2(x−1))) +(1/4)ln(x^2 +1)+c I = ∫_2 ^3 F(x)dx =[−(1/2)ln∣x−1∣]_2 ^3 +(1/4)[ln(x^2 +1)]_2 ^3 −(5/2)[ (1/(x−1))]_2 ^3 =−(1/2)ln(2) +(1/4){ln(10)−ln(5)} −(5/2){ −(1/2)} =−(1/2)ln(2) +(1/4)ln(2) +(5/4) =−(1/4)ln(2) +(5/4) .$
$l e t ? d e c o m p o s e$ $F (x) = \frac{2 x^{2} + 3}{{(x - 1)}^{2} (x^{2} + 1)}$ $F (x) = \frac{a}{x - 1} + \frac{b}{{(x - 1)}^{2}} + \frac{c x + d}{x^{2} + 1}$ $b = {l i m}_{x \to 1} {(x - 1)}^{2} F (x) = \frac{5}{2}$ ${l i m}_{x \to + \infty} x F (x) = 0 = a + c \Rightarrow c = - a \Rightarrow$ $F (x) = \frac{a}{x - 1} + \frac{5}{2 {(x - 1)}^{2}} + \frac{- a x + d}{x^{2} + 1}$ $F (o) = 3 = - a + \frac{5}{2} + d \Rightarrow d = 3 + a - \frac{5}{2} \Rightarrow$ $F (x) = \frac{a}{x - 1} + \frac{5}{2 {(x - 1)}^{2}} + \frac{- a x + a + \frac{1}{2}}{x^{2} + 1}$ $F (2) = a + \frac{5}{2} + \frac{- a + \frac{1}{2}}{5} = \frac{11}{5} \Rightarrow$ $5 a + \frac{25}{2} - a + \frac{1}{2} = 11 \Rightarrow 4 a + 13 = 11 \Rightarrow$ $4 a = - 2 \Rightarrow a = - \frac{1}{2} \Rightarrow c = \frac{1}{2} a n d d = 3 - \frac{1}{2} - \frac{5}{2} = 0$ $F (x) = \frac{- 1}{2 (x - 1)} + \frac{5}{2 {(x - 1)}^{2}} + \frac{\frac{1}{2} x}{x^{2} + 1}$ $\int F (x) d x = - \frac{1}{2} l n ∣ x - 1 ∣ - \frac{5}{2 (x - 1)} + \frac{1}{4} l n (x^{2} + 1) + c$ $I = \int_{2}^{3} F (x) d x = {[- \frac{1}{2} l n ∣ x - 1 ∣]}_{2}^{3} + \frac{1}{4} {[l n (x^{2} + 1)]}_{2}^{3}$ $- \frac{5}{2} {[\frac{1}{x - 1}]}_{2}^{3}$ $= - \frac{1}{2} l n (2) + \frac{1}{4} {l n (10) - l n (5)} - \frac{5}{2} {- \frac{1}{2}}$ $= - \frac{1}{2} l n (2) + \frac{1}{4} l n (2) + \frac{5}{4} = - \frac{1}{4} l n (2) + \frac{5}{4} .$
	Answered by ajfour last updated on 13/May/18

	$\frac{2 x^{2} + 3}{{(x - 1)}^{2} (x^{2} + 1)} = \frac{A}{x - 1} + \frac{B}{{(x - 1)}^{2}} + \frac{C x + d}{x^{2} + 1}$ $2 x^{2} + 3 = A (x^{2} + 1) (x - 1) + B (x^{2} + 1)$ $+ (C x + D) {(x - 1)}^{2}$ $= (A + C) x^{3} + (- A + B - 2 C + D) x^{2}$ $(A + C + 2 D) x - A + B + D$ $\Rightarrow C = - A; - A + B + D = 3$ $3 - 2 C = 2 o r C = 1 / 2$ $A = - \frac{1}{2}; D = 0, B = 3 + A$ $s o B = \frac{5}{2}$ $I = - \frac{1}{2} \ln (x - 1) ∣_{2}^{3} - \frac{5}{2 (x - 1)} ∣_{2}^{3}$ $+ \frac{1}{4} \ln (x^{2} + 1) ∣_{2}^{3}$ $I = - \frac{1}{2} \ln 2 + \frac{5}{4} + \frac{1}{4} \ln 2$ $I = \frac{5 - \ln 2}{4} .$
	Commented by math khazana by abdo last updated on 13/May/18

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