Question Number 34911 by abdo imad last updated on 12/May/18
$${find}\:\:\:\int_{\mathrm{2}} ^{\mathrm{3}} \:\:\:\:\frac{\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{3}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}\:{dx} \\ $$
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) .](Q34939.png)
$${let}?{decompose}\: \\ $$$${F}\left({x}\right)=\:\frac{\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{3}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$${F}\left({x}\right)\:=\:\frac{{a}}{{x}−\mathrm{1}}\:+\frac{{b}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{cx}+{d}}{{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${b}={lim}_{{x}\rightarrow\mathrm{1}} \left({x}−\mathrm{1}\right)^{\mathrm{2}} {F}\left({x}\right)=\:\frac{\mathrm{5}}{\mathrm{2}} \\ $$$${lim}_{{x}\rightarrow+\infty} \:{xF}\left({x}\right)=\:\mathrm{0}=\:{a}\:+{c}\:\Rightarrow{c}=−{a}\:\Rightarrow \\ $$$${F}\left({x}\right)=\:\frac{{a}}{{x}−\mathrm{1}}\:\:+\frac{\mathrm{5}}{\mathrm{2}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{−{ax}\:+{d}}{{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${F}\left({o}\right)\:=\mathrm{3}\:=\:−{a}\:\:+\frac{\mathrm{5}}{\mathrm{2}}\:\:+{d}\:\Rightarrow{d}\:=\mathrm{3}\:+{a}−\frac{\mathrm{5}}{\mathrm{2}}\:\Rightarrow \\ $$$${F}\left({x}\right)=\:\:\frac{{a}}{{x}−\mathrm{1}}\:+\frac{\mathrm{5}}{\mathrm{2}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:\:+\frac{−{ax}\:\:+{a}\:+\frac{\mathrm{1}}{\mathrm{2}}}{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$${F}\left(\mathrm{2}\right)\:=\:{a}\:+\frac{\mathrm{5}}{\mathrm{2}}\:\:+\frac{−{a}\:+\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{5}}\:=\:\frac{\mathrm{11}}{\mathrm{5}}\:\Rightarrow \\ $$$$\mathrm{5}{a}\:\:+\frac{\mathrm{25}}{\mathrm{2}}\:\:−{a}\:+\frac{\mathrm{1}}{\mathrm{2}}\:=\mathrm{11}\:\Rightarrow\mathrm{4}{a}\:\:+\mathrm{13}=\mathrm{11}\:\Rightarrow \\ $$$$\mathrm{4}{a}\:=\:−\mathrm{2}\:\Rightarrow{a}\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\:\Rightarrow{c}=\frac{\mathrm{1}}{\mathrm{2}}\:\:{and}\:\:{d}\:=\mathrm{3}−\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{\mathrm{5}}{\mathrm{2}}=\mathrm{0} \\ $$$${F}\left({x}\right)\:=\:\frac{−\mathrm{1}}{\mathrm{2}\left({x}−\mathrm{1}\right)}\:\:+\frac{\mathrm{5}}{\mathrm{2}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:\:+\:\frac{\frac{\mathrm{1}}{\mathrm{2}}{x}}{{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\int\:{F}\left({x}\right){dx}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{x}−\mathrm{1}\mid\:\:−\frac{\mathrm{5}}{\mathrm{2}\left({x}−\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{4}}{ln}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)+{c} \\ $$$${I}\:=\:\int_{\mathrm{2}} ^{\mathrm{3}} \:{F}\left({x}\right){dx}\:=\left[−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{x}−\mathrm{1}\mid\right]_{\mathrm{2}} ^{\mathrm{3}} \:\:+\frac{\mathrm{1}}{\mathrm{4}}\left[{ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)\right]_{\mathrm{2}} ^{\mathrm{3}} \\ $$$$−\frac{\mathrm{5}}{\mathrm{2}}\left[\:\frac{\mathrm{1}}{{x}−\mathrm{1}}\right]_{\mathrm{2}} ^{\mathrm{3}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:+\frac{\mathrm{1}}{\mathrm{4}}\left\{{ln}\left(\mathrm{10}\right)−{ln}\left(\mathrm{5}\right)\right\}\:−\frac{\mathrm{5}}{\mathrm{2}}\left\{\:−\frac{\mathrm{1}}{\mathrm{2}}\right\} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:\:+\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\mathrm{2}\right)\:+\frac{\mathrm{5}}{\mathrm{4}}\:=−\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\mathrm{2}\right)\:+\frac{\mathrm{5}}{\mathrm{4}}\:. \\ $$
Answered by ajfour last updated on 13/May/18
$$\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{{A}}{{x}−\mathrm{1}}+\frac{{B}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }+\frac{{Cx}+{d}}{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}={A}\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}−\mathrm{1}\right)+{B}\left({x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left({Cx}+{D}\right)\left({x}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\:=\left({A}+{C}\right){x}^{\mathrm{3}} +\left(−{A}+{B}−\mathrm{2}{C}+{D}\right){x}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\left({A}+{C}+\mathrm{2}{D}\right){x}−{A}+{B}+{D} \\ $$$$\Rightarrow\:\:{C}=−{A}\:\:;\:\:−{A}+{B}+{D}=\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\mathrm{3}−\mathrm{2}{C}=\mathrm{2}\:\:\:{or}\:\:\:\:{C}=\mathrm{1}/\mathrm{2} \\ $$$$\:\:\:\:{A}=−\frac{\mathrm{1}}{\mathrm{2}}\:\:;\:\:{D}=\mathrm{0},\:\:{B}=\mathrm{3}+{A} \\ $$$${so}\:\:{B}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$${I}=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({x}−\mathrm{1}\right)\mid_{\mathrm{2}} ^{\mathrm{3}} −\frac{\mathrm{5}}{\mathrm{2}\left({x}−\mathrm{1}\right)}\mid_{\mathrm{2}} ^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)\mid_{\mathrm{2}} ^{\mathrm{3}} \\ $$$${I}=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mathrm{2}+\frac{\mathrm{5}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mathrm{2} \\ $$$$\:\:\:\:{I}=\frac{\mathrm{5}−\mathrm{ln}\:\mathrm{2}}{\mathrm{4}}\:\:. \\ $$
Commented by math khazana by abdo last updated on 13/May/18
$${your}\:{answer}\:{is}\:{correct}\:{sir}\:{Ajfour}\:{thanks}... \\ $$