Question Number 57224 by maxmathsup by imad last updated on 31/Mar/19
$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{5}{t}\:+\mathrm{1}}{\left({t}+\mathrm{1}\right)\left({t}+\mathrm{2}\right)\left(\mathrm{2}{t}+\mathrm{3}\right)}{dt} \\ $$
Commented by maxmathsup by imad last updated on 01/Apr/19
![let decompose F(t)=((3t^2 −5t +1)/((t+1)(t+2)(2t+3))) with I=∫_0 ^1 F(t)dt F(t)=(a/(t+1)) +(b/(t+2)) +(c/(2t+3)) a =lim_(t→−1) (t+1)F(t) =(9/1) =9 b =lim_(t→−2) (t+2)F(t) =((12+10+1)/((−1)(−1))) =23 c =lim_(t→−(3/2)) (2t+3)F(t)=((3(9/4)−5(−(3/2))+1)/((−(3/2)+1)(−(3/2)+2))) =((((27)/4) +((30)/4) +1)/((−(1/2))((1/2)))) =((61)/4)(−4)=−61 ⇒ F(t)=(9/(t+1)) +((23)/(t+2)) −((61)/(2t+3)) ⇒∫_0 ^1 F(t)dt =[9ln∣t+1∣+23ln∣t+2∣−((61)/2)ln∣2t+3∣]_0 ^1 =9ln(2)+23ln(3)−((61)/2)ln(5) −23ln(2)+((61)/2)ln(3) I=−14ln(2) +((107)/2)ln(3)−((61)/2)ln(5) .](https://www.tinkutara.com/question/Q57301.png)
$${let}\:{decompose}\:{F}\left({t}\right)=\frac{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{5}{t}\:+\mathrm{1}}{\left({t}+\mathrm{1}\right)\left({t}+\mathrm{2}\right)\left(\mathrm{2}{t}+\mathrm{3}\right)}\:\:{with}\:{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \:{F}\left({t}\right){dt} \\ $$$${F}\left({t}\right)=\frac{{a}}{{t}+\mathrm{1}}\:+\frac{{b}}{{t}+\mathrm{2}}\:+\frac{{c}}{\mathrm{2}{t}+\mathrm{3}} \\ $$$${a}\:={lim}_{{t}\rightarrow−\mathrm{1}} \left({t}+\mathrm{1}\right){F}\left({t}\right)\:=\frac{\mathrm{9}}{\mathrm{1}}\:=\mathrm{9} \\ $$$${b}\:={lim}_{{t}\rightarrow−\mathrm{2}} \left({t}+\mathrm{2}\right){F}\left({t}\right)\:=\frac{\mathrm{12}+\mathrm{10}+\mathrm{1}}{\left(−\mathrm{1}\right)\left(−\mathrm{1}\right)}\:=\mathrm{23} \\ $$$${c}\:={lim}_{{t}\rightarrow−\frac{\mathrm{3}}{\mathrm{2}}} \:\:\left(\mathrm{2}{t}+\mathrm{3}\right){F}\left({t}\right)=\frac{\mathrm{3}\frac{\mathrm{9}}{\mathrm{4}}−\mathrm{5}\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)+\mathrm{1}}{\left(−\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{1}\right)\left(−\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{2}\right)}\:=\frac{\frac{\mathrm{27}}{\mathrm{4}}\:+\frac{\mathrm{30}}{\mathrm{4}}\:+\mathrm{1}}{\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}\:=\frac{\mathrm{61}}{\mathrm{4}}\left(−\mathrm{4}\right)=−\mathrm{61}\:\Rightarrow \\ $$$${F}\left({t}\right)=\frac{\mathrm{9}}{{t}+\mathrm{1}}\:+\frac{\mathrm{23}}{{t}+\mathrm{2}}\:−\frac{\mathrm{61}}{\mathrm{2}{t}+\mathrm{3}}\:\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} {F}\left({t}\right){dt}\:=\left[\mathrm{9}{ln}\mid{t}+\mathrm{1}\mid+\mathrm{23}{ln}\mid{t}+\mathrm{2}\mid−\frac{\mathrm{61}}{\mathrm{2}}{ln}\mid\mathrm{2}{t}+\mathrm{3}\mid\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\mathrm{9}{ln}\left(\mathrm{2}\right)+\mathrm{23}{ln}\left(\mathrm{3}\right)−\frac{\mathrm{61}}{\mathrm{2}}{ln}\left(\mathrm{5}\right)\:−\mathrm{23}{ln}\left(\mathrm{2}\right)+\frac{\mathrm{61}}{\mathrm{2}}{ln}\left(\mathrm{3}\right) \\ $$$${I}=−\mathrm{14}{ln}\left(\mathrm{2}\right)\:+\frac{\mathrm{107}}{\mathrm{2}}{ln}\left(\mathrm{3}\right)−\frac{\mathrm{61}}{\mathrm{2}}{ln}\left(\mathrm{5}\right)\:. \\ $$
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) ))}](https://www.tinkutara.com/question/Q57243.png)
$$\frac{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{5}{t}+\mathrm{1}}{\left({t}+\mathrm{1}\right)\left({t}+\mathrm{2}\right)\left(\mathrm{2}{t}+\mathrm{3}\right)}=\frac{{a}}{{t}+\mathrm{1}}+\frac{{b}}{{t}+\mathrm{2}}+\frac{{c}}{\mathrm{2}{t}+\mathrm{3}} \\ $$$$\mathrm{3}{t}^{\mathrm{2}} −\mathrm{5}{t}+\mathrm{1}={a}\left({t}+\mathrm{2}\right)\left(\mathrm{2}{t}+\mathrm{3}\right)+{b}\left({t}+\mathrm{1}\right)\left(\mathrm{2}{t}+\mathrm{3}\right)+{c}\left({t}+\mathrm{1}\right)\left({t}+\mathrm{2}\right) \\ $$$${put}\:{t}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{3}+\mathrm{5}+\mathrm{1}={a}\left(−\mathrm{1}+\mathrm{2}\right)\left(−\mathrm{2}+\mathrm{3}\right) \\ $$$${a}=\mathrm{9} \\ $$$${put}\:{t}+\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{3}\left(\mathrm{4}\right)+\mathrm{10}+\mathrm{1}={b}\left(−\mathrm{2}+\mathrm{1}\right)\left(−\mathrm{4}+\mathrm{3}\right) \\ $$$${b}=\mathrm{23} \\ $$$${put}\:\mathrm{2}{t}+\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{3}\left(\frac{−\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{5}\left(\frac{−\mathrm{3}}{\mathrm{2}}\right)+\mathrm{1}={c}\left(\frac{−\mathrm{3}}{\mathrm{2}}+\mathrm{1}\right)\left(\frac{−\mathrm{3}}{\mathrm{2}}+\mathrm{2}\right) \\ $$$$\mathrm{3}\left(\frac{\mathrm{9}}{\mathrm{4}}\right)+\frac{\mathrm{15}}{\mathrm{2}}+\mathrm{1}={c}\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{27}+\mathrm{30}}{\mathrm{4}}\:+\mathrm{1}=\frac{−\mathrm{1}}{\mathrm{4}}×{c} \\ $$$$\frac{\mathrm{61}}{\mathrm{4}}=\frac{−{c}}{\mathrm{4}}\rightarrow{c}=−\mathrm{61} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{a}}{{t}+\mathrm{1}}{dt}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{b}}{{t}+\mathrm{2}}{dt}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{c}}{\mathrm{2}{t}+\mathrm{3}}{dt} \\ $$$$\mathrm{9}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{{t}+\mathrm{1}}+\mathrm{23}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{{t}+\mathrm{2}}−\mathrm{61}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{{t}+\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$=\mid\mathrm{9}{ln}\left({t}+\mathrm{1}\right)+\mathrm{23}{ln}\left({t}+\mathrm{2}\right)−\mathrm{61}{ln}\left({t}+\frac{\mathrm{3}}{\mathrm{2}}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\left[\mathrm{9}\left\{{ln}\left(\mathrm{2}\right)−{ln}\left(\mathrm{1}\right)\right\}+\mathrm{23}\left\{{ln}\left(\mathrm{3}\right)−{ln}\mathrm{2}\right\}−\mathrm{61}\left\{{ln}\left(\frac{\mathrm{5}}{\mathrm{2}}\right)−{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\right\}\right] \\ $$$$=\mathrm{9}{ln}\mathrm{2}+\mathrm{23}{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)−\mathrm{61}{ln}\left(\frac{\mathrm{5}}{\mathrm{3}}\right) \\ $$$$={ln}\mathrm{2}^{\mathrm{9}} +{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{23}} −{ln}\left(\frac{\mathrm{5}}{\mathrm{3}}\right)^{\mathrm{61}} \\ $$$${ln}\left\{\mathrm{2}^{\mathrm{9}} ×\frac{\mathrm{3}^{\mathrm{23}} }{\mathrm{2}^{\mathrm{23}} }×\frac{\mathrm{3}^{\mathrm{61}} }{\mathrm{5}^{\mathrm{61}} }\right\} \\ $$$$={ln}\left\{\frac{\mathrm{3}^{\mathrm{84}} }{\mathrm{2}^{\mathrm{14}} ×\mathrm{5}^{\mathrm{61}} }\right\}\: \\ $$