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Question Number 48643 by cesar.marval.larez@gmail.com last updated on 26/Nov/18

Commented by maxmathsup by imad last updated on 26/Nov/18

1)let I =∫ ((3x+2)/((x−1)(x−2)(x^2 +3)))dx let decompose F(x)=((3x+2)/((x−1)(x−2)(x^2 +3))) F(x) =(a/(x−1)) +(b/(x−2)) +((cx+d)/(x^(2 ) +3)) a =lim_(x→1) (x−1)F(x) =−(5/4) b =lim_(x→2) (x−2)F(x) =(8/7) lim_(x→+∞) xF(x) =0 =a+b +c ⇒c=(5/4) −(8/7) =((35−32)/(28)) =(3/(28)) F(0) =(1/3) =−a−(b/2) +(d/3) =(5/4) −(4/7) +(d/3) ⇒1 =((15)/4) −((12)/7) +d ⇒ 1=((105−48)/(28)) +d =((57)/(28)) +d ⇒d=1−((57)/(28)) =−((57−28)/(28)) =−((99)/(28)) ⇒ F(x) =−(5/(4(x−1))) +(8/(7(x−2))) +(((3/(28 ))x−((99)/(28)))/(x^2 +3)) ⇒I =∫ F(x)dx =−(5/4)ln∣x−1∣ +(8/7)ln∣x−2∣ +(3/(56))∫ ((2x)/(x^2 +3))dx −((99)/(28)) ∫ (dx/(x^2 +3)) =−(5/4)ln∣x−1∣ +(8/7)ln∣x−2∣ +(3/(56))ln(x^2 +3) −((99)/(28)) ∫ (dx/(x^2 +3)) but ∫ (dx/(x^2 +3)) =_(x=(√3)t) ∫ (1/(3(1+t^2 )))(√3)dt =(1/(√3)) arctan((x/(√3))) +c ⇒ I =−(5/4)ln∣x−1∣+(8/7)ln∣x−2∣ +(3/(56))ln(x^2 +3)−((99)/(28(√3))) arctan((x/(√3)))+C .

$$\left.\mathrm{1}\right){let}\:\:{I}\:=\int\:\:\frac{\mathrm{3}{x}+\mathrm{2}}{\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} \:+\mathrm{3}\right)}{dx}\:{let}\:{decompose}\:{F}\left({x}\right)=\frac{\mathrm{3}{x}+\mathrm{2}}{\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} \:+\mathrm{3}\right)} \\ $$$${F}\left({x}\right)\:=\frac{{a}}{{x}−\mathrm{1}}\:+\frac{{b}}{{x}−\mathrm{2}}\:+\frac{{cx}+{d}}{{x}^{\mathrm{2}\:} +\mathrm{3}} \\ $$$${a}\:={lim}_{{x}\rightarrow\mathrm{1}} \left({x}−\mathrm{1}\right){F}\left({x}\right)\:=−\frac{\mathrm{5}}{\mathrm{4}} \\ $$$${b}\:={lim}_{{x}\rightarrow\mathrm{2}} \left({x}−\mathrm{2}\right){F}\left({x}\right)\:=\frac{\mathrm{8}}{\mathrm{7}} \\ $$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)\:=\mathrm{0}\:={a}+{b}\:+{c}\:\Rightarrow{c}=\frac{\mathrm{5}}{\mathrm{4}}\:−\frac{\mathrm{8}}{\mathrm{7}}\:=\frac{\mathrm{35}−\mathrm{32}}{\mathrm{28}}\:=\frac{\mathrm{3}}{\mathrm{28}} \\ $$$${F}\left(\mathrm{0}\right)\:=\frac{\mathrm{1}}{\mathrm{3}}\:=−{a}−\frac{{b}}{\mathrm{2}}\:+\frac{{d}}{\mathrm{3}}\:=\frac{\mathrm{5}}{\mathrm{4}}\:−\frac{\mathrm{4}}{\mathrm{7}}\:+\frac{{d}}{\mathrm{3}}\:\Rightarrow\mathrm{1}\:=\frac{\mathrm{15}}{\mathrm{4}}\:−\frac{\mathrm{12}}{\mathrm{7}}\:+{d}\:\Rightarrow \\ $$$$\mathrm{1}=\frac{\mathrm{105}−\mathrm{48}}{\mathrm{28}}\:\:+{d}\:=\frac{\mathrm{57}}{\mathrm{28}}\:+{d}\:\Rightarrow{d}=\mathrm{1}−\frac{\mathrm{57}}{\mathrm{28}}\:=−\frac{\mathrm{57}−\mathrm{28}}{\mathrm{28}}\:=−\frac{\mathrm{99}}{\mathrm{28}}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=−\frac{\mathrm{5}}{\mathrm{4}\left({x}−\mathrm{1}\right)}\:+\frac{\mathrm{8}}{\mathrm{7}\left({x}−\mathrm{2}\right)}\:+\frac{\frac{\mathrm{3}}{\mathrm{28}\:}{x}−\frac{\mathrm{99}}{\mathrm{28}}}{{x}^{\mathrm{2}} \:+\mathrm{3}}\:\Rightarrow{I}\:=\int\:{F}\left({x}\right){dx} \\ $$$$=−\frac{\mathrm{5}}{\mathrm{4}}{ln}\mid{x}−\mathrm{1}\mid\:+\frac{\mathrm{8}}{\mathrm{7}}{ln}\mid{x}−\mathrm{2}\mid\:\:+\frac{\mathrm{3}}{\mathrm{56}}\int\:\:\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} \:+\mathrm{3}}{dx}\:−\frac{\mathrm{99}}{\mathrm{28}}\:\int\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{3}} \\ $$$$=−\frac{\mathrm{5}}{\mathrm{4}}{ln}\mid{x}−\mathrm{1}\mid\:+\frac{\mathrm{8}}{\mathrm{7}}{ln}\mid{x}−\mathrm{2}\mid\:+\frac{\mathrm{3}}{\mathrm{56}}{ln}\left({x}^{\mathrm{2}} \:+\mathrm{3}\right)\:−\frac{\mathrm{99}}{\mathrm{28}}\:\int\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{3}}\:{but} \\ $$$$\int\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{3}}\:=_{{x}=\sqrt{\mathrm{3}}{t}} \:\int\:\:\:\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\sqrt{\mathrm{3}}{dt}\:=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:{arctan}\left(\frac{{x}}{\sqrt{\mathrm{3}}}\right)\:+{c}\:\Rightarrow \\ $$$${I}\:=−\frac{\mathrm{5}}{\mathrm{4}}{ln}\mid{x}−\mathrm{1}\mid+\frac{\mathrm{8}}{\mathrm{7}}{ln}\mid{x}−\mathrm{2}\mid\:+\frac{\mathrm{3}}{\mathrm{56}}{ln}\left({x}^{\mathrm{2}} \:+\mathrm{3}\right)−\frac{\mathrm{99}}{\mathrm{28}\sqrt{\mathrm{3}}}\:{arctan}\left(\frac{{x}}{\sqrt{\mathrm{3}}}\right)+{C}\:. \\ $$$$ \\ $$

Commented by maxmathsup by imad last updated on 26/Nov/18

sorry error of typing d=−((29)/(28)) ⇒ I =−(5/4)ln∣x−1∣ +(8/7)ln∣x−2∣ +(3/(56))ln(x^2 +3)−((29)/(28(√3))) arctan((x/(√3)))+C .

$${sorry}\:{error}\:{of}\:{typing}\:\:\:{d}=−\frac{\mathrm{29}}{\mathrm{28}}\:\Rightarrow \\ $$$${I}\:=−\frac{\mathrm{5}}{\mathrm{4}}{ln}\mid{x}−\mathrm{1}\mid\:+\frac{\mathrm{8}}{\mathrm{7}}{ln}\mid{x}−\mathrm{2}\mid\:+\frac{\mathrm{3}}{\mathrm{56}}{ln}\left({x}^{\mathrm{2}} \:+\mathrm{3}\right)−\frac{\mathrm{29}}{\mathrm{28}\sqrt{\mathrm{3}}}\:{arctan}\left(\frac{{x}}{\sqrt{\mathrm{3}}}\right)+{C}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 26/Nov/18

2)∫((5x+3)/(x^2 +2))dx (5/2)∫((2x)/(x^2 +2))dx+3∫(dx/(x^2 +2)) (5/2)ln(x^2 +2)+(3/(√2))tan^(−1) ((x/(√2)))+c

$$\left.\mathrm{2}\right)\int\frac{\mathrm{5}{x}+\mathrm{3}}{{x}^{\mathrm{2}} +\mathrm{2}}{dx} \\ $$$$\frac{\mathrm{5}}{\mathrm{2}}\int\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{2}}{dx}+\mathrm{3}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{2}} \\ $$$$\frac{\mathrm{5}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} +\mathrm{2}\right)+\frac{\mathrm{3}}{\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\frac{{x}}{\sqrt{\mathrm{2}}}\right)+{c} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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