find the value of ∫_0 ^1 ((3t^2 −5t +1)/((t+1)(t+2)(2t+3)))dt


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Question Number 57224 by maxmathsup by imad last updated on 31/Mar/19

$f i n d t h e v a l u e o f \int_{0}^{1} \frac{3 t^{2} - 5 t + 1}{(t + 1) (t + 2) (2 t + 3)} d t$
Commented by maxmathsup by imad last updated on 01/Apr/19

$l e t d e c o m p o s e F (t) = \frac{3 t^{2} - 5 t + 1}{(t + 1) (t + 2) (2 t + 3)} w i t h I = \int_{0}^{1} F (t) d t$ $F (t) = \frac{a}{t + 1} + \frac{b}{t + 2} + \frac{c}{2 t + 3}$ $a = {l i m}_{t \to - 1} (t + 1) F (t) = \frac{9}{1} = 9$ $b = {l i m}_{t \to - 2} (t + 2) F (t) = \frac{12 + 10 + 1}{(- 1) (- 1)} = 23$ $c = {l i m}_{t \to - \frac{3}{2}} (2 t + 3) F (t) = \frac{3 \frac{9}{4} - 5 (- \frac{3}{2}) + 1}{(- \frac{3}{2} + 1) (- \frac{3}{2} + 2)} = \frac{\frac{27}{4} + \frac{30}{4} + 1}{(- \frac{1}{2}) (\frac{1}{2})} = \frac{61}{4} (- 4) = - 61 \Rightarrow$ $F (t) = \frac{9}{t + 1} + \frac{23}{t + 2} - \frac{61}{2 t + 3} \Rightarrow \int_{0}^{1} F (t) d t = {[9 l n ∣ t + 1 ∣ + 23 l n ∣ t + 2 ∣ - \frac{61}{2} l n ∣ 2 t + 3 ∣]}_{0}^{1}$ $= 9 l n (2) + 23 l n (3) - \frac{61}{2} l n (5) - 23 l n (2) + \frac{61}{2} l n (3)$ $I = - 14 l n (2) + \frac{107}{2} l n (3) - \frac{61}{2} l n (5) .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 01/Apr/19
	$((3t^2 −5t+1)/((t+1)(t+2)(2t+3)))=(a/(t+1))+(b/(t+2))+(c/(2t+3)) 3t^2 −5t+1=a(t+2)(2t+3)+b(t+1)(2t+3)+c(t+1)(t+2) put t+1=0 3+5+1=a(−1+2)(−2+3) a=9 put t+2=0 3(4)+10+1=b(−2+1)(−4+3) b=23 put 2t+3=0 3(((−3)/2))^2 −5(((−3)/2))+1=c(((−3)/2)+1)(((−3)/2)+2) 3((9/4))+((15)/2)+1=c(((−1)/2))((1/2)) ((27+30)/4) +1=((−1)/4)×c ((61)/4)=((−c)/4)→c=−61 ∫_0 ^1 (a/(t+1))dt+∫_0 ^1 (b/(t+2))dt+∫_0 ^1 (c/(2t+3))dt 9∫_0 ^1 (dt/(t+1))+23∫_0 ^1 (dt/(t+2))−61∫_0 ^1 (dt/(t+(3/2))) =∣9ln(t+1)+23ln(t+2)−61ln(t+(3/2))∣_0 ^1 =[9{ln(2)−ln(1)}+23{ln(3)−ln2}−61{ln((5/2))−ln((3/2))}] =9ln2+23ln((3/2))−61ln((5/3)) =ln2^9 +ln((3/2))^(23) −ln((5/3))^(61) ln{2^9 ×(3^(23) /2^(23) )×(3^(61) /5^(61) )} =ln{(3^(84) /(2^(14) ×5^(61) ))}$
	$\frac{3 t^{2} - 5 t + 1}{(t + 1) (t + 2) (2 t + 3)} = \frac{a}{t + 1} + \frac{b}{t + 2} + \frac{c}{2 t + 3}$ $3 t^{2} - 5 t + 1 = a (t + 2) (2 t + 3) + b (t + 1) (2 t + 3) + c (t + 1) (t + 2)$ $p u t t + 1 = 0$ $3 + 5 + 1 = a (- 1 + 2) (- 2 + 3)$ $a = 9$ $p u t t + 2 = 0$ $3 (4) + 10 + 1 = b (- 2 + 1) (- 4 + 3)$ $b = 23$ $p u t 2 t + 3 = 0$ $3 {(\frac{- 3}{2})}^{2} - 5 (\frac{- 3}{2}) + 1 = c (\frac{- 3}{2} + 1) (\frac{- 3}{2} + 2)$ $3 (\frac{9}{4}) + \frac{15}{2} + 1 = c (\frac{- 1}{2}) (\frac{1}{2})$ $\frac{27 + 30}{4} + 1 = \frac{- 1}{4} \times c$ $\frac{61}{4} = \frac{- c}{4} \to c = - 61$ $\int_{0}^{1} \frac{a}{t + 1} d t + \int_{0}^{1} \frac{b}{t + 2} d t + \int_{0}^{1} \frac{c}{2 t + 3} d t$ $9 \int_{0}^{1} \frac{d t}{t + 1} + 23 \int_{0}^{1} \frac{d t}{t + 2} - 61 \int_{0}^{1} \frac{d t}{t + \frac{3}{2}}$ $=∣ 9 l n (t + 1) + 23 l n (t + 2) - 61 l n (t + \frac{3}{2}) ∣_{0}^{1}$ $= [9 {l n (2) - l n (1)} + 23 {l n (3) - l n 2} - 61 {l n (\frac{5}{2}) - l n (\frac{3}{2})}]$ $= 9 l n 2 + 23 l n (\frac{3}{2}) - 61 l n (\frac{5}{3})$ $= l n 2^{9} + l n {(\frac{3}{2})}^{23} - l n {(\frac{5}{3})}^{61}$ $l n {2^{9} \times \frac{3^{23}}{2^{23}} \times \frac{3^{61}}{5^{61}}}$ $= l n {\frac{3^{84}}{2^{14} \times 5^{61}}}$
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