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Question-85896

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Question Number 85896 by ar247 last updated on 25/Mar/20
Commented by abdomathmax last updated on 25/Mar/20
I=∫ ((5x−5)/(3x^2 −8x−3))dx 3x^2 −8x−3=0 →Δ^′ =4^2 −(3)(−3) =16+9=25 x_1 =((4+5)/3) =3 and x_2 =((4−5)/3)=−(1/3) ⇒ I =∫ ((5x−5)/(3(x−3)(x+(1/3))))dx let decompose F(x)=((5x−5)/((x−3)(3x+1))) F(x)=(a/(x−3)) +(b/(3x+1)) a=(x−3)F(x)∣_(x=3) =((10)/(10))=1 b=(3x+1)F(x)∣_(x=−(1/3)) = ((−(5/3)−5)/(−(1/3)−3)) =((20)/(10))=2 ⇒ F(x)=(1/(x−3))+(2/(3x+1)) ⇒ I =∫(dx/(x−3)) +2∫ (dx/(3x+1)) =ln∣x−3∣+(2/3)ln∣3x+1∣ +c
$${I}=\int\:\:\frac{\mathrm{5}{x}−\mathrm{5}}{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{8}{x}−\mathrm{3}}{dx} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} −\mathrm{8}{x}−\mathrm{3}=\mathrm{0}\:\rightarrow\Delta^{'} =\mathrm{4}^{\mathrm{2}} −\left(\mathrm{3}\right)\left(−\mathrm{3}\right)\:=\mathrm{16}+\mathrm{9}=\mathrm{25} \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{4}+\mathrm{5}}{\mathrm{3}}\:=\mathrm{3}\:{and}\:{x}_{\mathrm{2}} =\frac{\mathrm{4}−\mathrm{5}}{\mathrm{3}}=−\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow \\ $$$${I}\:=\int\:\:\:\frac{\mathrm{5}{x}−\mathrm{5}}{\mathrm{3}\left({x}−\mathrm{3}\right)\left({x}+\frac{\mathrm{1}}{\mathrm{3}}\right)}{dx}\:{let}\:{decompose} \\ $$$${F}\left({x}\right)=\frac{\mathrm{5}{x}−\mathrm{5}}{\left({x}−\mathrm{3}\right)\left(\mathrm{3}{x}+\mathrm{1}\right)} \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}−\mathrm{3}}\:+\frac{{b}}{\mathrm{3}{x}+\mathrm{1}} \\ $$$${a}=\left({x}−\mathrm{3}\right){F}\left({x}\right)\mid_{{x}=\mathrm{3}} \:\:\:=\frac{\mathrm{10}}{\mathrm{10}}=\mathrm{1} \\ $$$${b}=\left(\mathrm{3}{x}+\mathrm{1}\right){F}\left({x}\right)\mid_{{x}=−\frac{\mathrm{1}}{\mathrm{3}}} \:\:\:=\:\frac{−\frac{\mathrm{5}}{\mathrm{3}}−\mathrm{5}}{−\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{3}}\:=\frac{\mathrm{20}}{\mathrm{10}}=\mathrm{2}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{{x}−\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}{x}+\mathrm{1}}\:\Rightarrow\:{I}\:=\int\frac{{dx}}{{x}−\mathrm{3}}\:+\mathrm{2}\int\:\:\frac{{dx}}{\mathrm{3}{x}+\mathrm{1}} \\ $$$$={ln}\mid{x}−\mathrm{3}\mid+\frac{\mathrm{2}}{\mathrm{3}}{ln}\mid\mathrm{3}{x}+\mathrm{1}\mid\:+{c} \\ $$
Commented by abdomathmax last updated on 25/Mar/20
2) A =∫ ((2x^5 −x^3 −1)/(x^3 −4x))dx ⇒ A =∫ ((2(x^3 −4x)x^2 +8x^3 −x^3 −1)/(x^3 −4x))dx =2x +∫ ((7x^3 −1)/(x^3 −4x))dx =2x +∫ ((7(x^3 −4x)+28x−1)/(x^3 −4x))dx =2x +7 +∫ ((28x−1)/(x^3 −4x))dx let decompose F(x)=((28x−1)/(x^3 −4x)) =((28x−1)/(x(x−2)(x+2))) F(x)=(a/x) +(b/(x−2)) +(c/(x+2)) a =((−1)/((−2)(2))) =(1/4) b=((56−1)/(2.4)) =((55)/8) c=((−56−1)/((−2)(−4))) =((57)/8) ⇒F(x)=(1/(4x))+((55)/(8(x−2)))+((57)/(8(x+2))) A =2x+7 +(1/4)ln∣x∣ +((55)/8)ln∣x−2∣ +((57)/8)ln∣x+2∣ +C
$$\left.\mathrm{2}\right)\:{A}\:=\int\:\:\frac{\mathrm{2}{x}^{\mathrm{5}} −{x}^{\mathrm{3}} −\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{4}{x}}{dx}\:\:\Rightarrow \\ $$$${A}\:=\int\:\:\frac{\mathrm{2}\left({x}^{\mathrm{3}} −\mathrm{4}{x}\right){x}^{\mathrm{2}} +\mathrm{8}{x}^{\mathrm{3}} −{x}^{\mathrm{3}} −\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{4}{x}}{dx} \\ $$$$=\mathrm{2}{x}\:\:+\int\:\:\:\frac{\mathrm{7}{x}^{\mathrm{3}} −\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{4}{x}}{dx} \\ $$$$=\mathrm{2}{x}\:+\int\:\:\frac{\mathrm{7}\left({x}^{\mathrm{3}} −\mathrm{4}{x}\right)+\mathrm{28}{x}−\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{4}{x}}{dx} \\ $$$$=\mathrm{2}{x}\:+\mathrm{7}\:+\int\:\:\:\frac{\mathrm{28}{x}−\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{4}{x}}{dx}\:{let}\:{decompose} \\ $$$${F}\left({x}\right)=\frac{\mathrm{28}{x}−\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{4}{x}}\:=\frac{\mathrm{28}{x}−\mathrm{1}}{{x}\left({x}−\mathrm{2}\right)\left({x}+\mathrm{2}\right)} \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}}\:+\frac{{b}}{{x}−\mathrm{2}}\:+\frac{{c}}{{x}+\mathrm{2}} \\ $$$${a}\:=\frac{−\mathrm{1}}{\left(−\mathrm{2}\right)\left(\mathrm{2}\right)}\:=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${b}=\frac{\mathrm{56}−\mathrm{1}}{\mathrm{2}.\mathrm{4}}\:=\frac{\mathrm{55}}{\mathrm{8}} \\ $$$${c}=\frac{−\mathrm{56}−\mathrm{1}}{\left(−\mathrm{2}\right)\left(−\mathrm{4}\right)}\:=\frac{\mathrm{57}}{\mathrm{8}}\:\Rightarrow{F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}{x}}+\frac{\mathrm{55}}{\mathrm{8}\left({x}−\mathrm{2}\right)}+\frac{\mathrm{57}}{\mathrm{8}\left({x}+\mathrm{2}\right)} \\ $$$${A}\:=\mathrm{2}{x}+\mathrm{7}\:+\frac{\mathrm{1}}{\mathrm{4}}{ln}\mid{x}\mid\:+\frac{\mathrm{55}}{\mathrm{8}}{ln}\mid{x}−\mathrm{2}\mid\:+\frac{\mathrm{57}}{\mathrm{8}}{ln}\mid{x}+\mathrm{2}\mid\:+{C} \\ $$$$ \\ $$
Commented by abdomathmax last updated on 25/Mar/20
4) I = ∫ ((√(x^2 −25))/x)dx we do the chsngement x=5 ch(t) ⇒ I =∫ ((5 sh(t))/(5ch(t))) ×5sh(t)dt =5 ∫ ((sh^2 t)/(cht)) dt =(5/2)∫ ((ch(2t)−1)/(ch(t)))dt =(5/2)∫ ((((e^(2t) +e^(−2t) )/2)−1)/((e^t +e^(−t) )/2))dt =(5/2)∫ ((e^(2t) +e^(−2t) −2)/(e^t +e^(−t) ))dt =_(e^t =u) (5/2)∫ ((u^2 +u^(−2) −2)/(u+u^(−1) ))(du/u) =(5/2)∫ ((u^2 +u^(−2) −2)/(u^2 +1))du =(5/2)∫ ((u^4 +1−2u^2 )/(u^2 (u^2 +1)))du decomposition ((u^4 −2u^2 +1)/(u^4 +u^2 )) =((u^4 +u^2 −3u^2 +1)/(u^4 +u^2 )) =1 +((−3u^2 +1)/(u^4 +u^2 )) F(u) =((−3u^2 +1)/(u^4 +u^2 )) =((−3u^2 +1)/(u^2 (u^2 +1))) =(a/u) +(b/u^2 ) +((cu +d)/(u^2 +1)) F(−u)=F(u) ⇒((−a)/u) +(b/u^2 ) +((−cu +d)/(u^2 +1))=(a/u) +(b/u^2 )+((cu+d)/(u^2 +1)) ⇒a=0 and c=0 ⇒F(u)=(b/u^2 ) +(d/(u^2 +1)) b=1 ⇒F(u)=(1/u^2 ) +(d/(u^2 +1)) F(1)=−1 =1 +(d/2) ⇒−2 =(d/2) ⇒d=−4 ⇒ F(u)=(1/u^2 )−(4/(u^2 +1)) ⇒ I =(5/2){u +∫ F(u)du} =(5/2)u +(5/2)(−(1/u)−4 arctan(u)) +C =(5/2)e^t −(5/2)e^(−t) −10 arctan(e^t )+C t=argch((x/5)) =ln((x/5)+(√((x^2 /(25))−1))) ⇒ I =(5/2){ (x/5)+(√((x^2 /(25))−1))−(1/((x/5)+(√((x^2 /(25))−1))))} −10 arctan((x/5)+(√((x^2 /(25))−1))) +C
$$\left.\mathrm{4}\right)\:{I}\:=\:\int\:\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{25}}}{{x}}{dx}\:\:{we}\:{do}\:{the}\:{chsngement} \\ $$$${x}=\mathrm{5}\:{ch}\left({t}\right)\:\Rightarrow\:{I}\:=\int\:\frac{\mathrm{5}\:{sh}\left({t}\right)}{\mathrm{5}{ch}\left({t}\right)}\:×\mathrm{5}{sh}\left({t}\right){dt} \\ $$$$=\mathrm{5}\:\int\:\:\frac{{sh}^{\mathrm{2}} {t}}{{cht}}\:{dt}\:=\frac{\mathrm{5}}{\mathrm{2}}\int\:\:\frac{{ch}\left(\mathrm{2}{t}\right)−\mathrm{1}}{{ch}\left({t}\right)}{dt} \\ $$$$=\frac{\mathrm{5}}{\mathrm{2}}\int\:\:\frac{\frac{{e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} }{\mathrm{2}}−\mathrm{1}}{\frac{{e}^{{t}} \:+{e}^{−{t}} }{\mathrm{2}}}{dt}\:=\frac{\mathrm{5}}{\mathrm{2}}\int\:\:\frac{{e}^{\mathrm{2}{t}} +{e}^{−\mathrm{2}{t}} −\mathrm{2}}{{e}^{{t}} \:+{e}^{−{t}} }{dt} \\ $$$$=_{{e}^{{t}} \:={u}} \:\:\:\:\frac{\mathrm{5}}{\mathrm{2}}\int\:\:\frac{{u}^{\mathrm{2}} \:+{u}^{−\mathrm{2}} −\mathrm{2}}{{u}+{u}^{−\mathrm{1}} }\frac{{du}}{{u}} \\ $$$$=\frac{\mathrm{5}}{\mathrm{2}}\int\:\:\frac{{u}^{\mathrm{2}} +{u}^{−\mathrm{2}} −\mathrm{2}}{{u}^{\mathrm{2}} \:+\mathrm{1}}{du} \\ $$$$=\frac{\mathrm{5}}{\mathrm{2}}\int\:\:\frac{{u}^{\mathrm{4}} \:+\mathrm{1}−\mathrm{2}{u}^{\mathrm{2}} }{{u}^{\mathrm{2}} \left({u}^{\mathrm{2}} \:+\mathrm{1}\right)}{du}\:\:{decomposition} \\ $$$$\frac{{u}^{\mathrm{4}} −\mathrm{2}{u}^{\mathrm{2}} \:+\mathrm{1}}{{u}^{\mathrm{4}} \:+{u}^{\mathrm{2}} }\:=\frac{{u}^{\mathrm{4}} +{u}^{\mathrm{2}} −\mathrm{3}{u}^{\mathrm{2}} +\mathrm{1}}{{u}^{\mathrm{4}} \:+{u}^{\mathrm{2}} } \\ $$$$=\mathrm{1}\:+\frac{−\mathrm{3}{u}^{\mathrm{2}} \:+\mathrm{1}}{{u}^{\mathrm{4}} \:+{u}^{\mathrm{2}} } \\ $$$${F}\left({u}\right)\:=\frac{−\mathrm{3}{u}^{\mathrm{2}} \:+\mathrm{1}}{{u}^{\mathrm{4}} \:+{u}^{\mathrm{2}} }\:=\frac{−\mathrm{3}{u}^{\mathrm{2}} +\mathrm{1}}{{u}^{\mathrm{2}} \left({u}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$$=\frac{{a}}{{u}}\:+\frac{{b}}{{u}^{\mathrm{2}} }\:+\frac{{cu}\:+{d}}{{u}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${F}\left(−{u}\right)={F}\left({u}\right)\:\Rightarrow\frac{−{a}}{{u}}\:+\frac{{b}}{{u}^{\mathrm{2}} }\:+\frac{−{cu}\:+{d}}{{u}^{\mathrm{2}} \:+\mathrm{1}}=\frac{{a}}{{u}}\:+\frac{{b}}{{u}^{\mathrm{2}} }+\frac{{cu}+{d}}{{u}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\Rightarrow{a}=\mathrm{0}\:\:{and}\:{c}=\mathrm{0}\:\Rightarrow{F}\left({u}\right)=\frac{{b}}{{u}^{\mathrm{2}} }\:+\frac{{d}}{{u}^{\mathrm{2}} +\mathrm{1}} \\ $$$${b}=\mathrm{1}\:\Rightarrow{F}\left({u}\right)=\frac{\mathrm{1}}{{u}^{\mathrm{2}} }\:+\frac{{d}}{{u}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${F}\left(\mathrm{1}\right)=−\mathrm{1}\:=\mathrm{1}\:+\frac{{d}}{\mathrm{2}}\:\Rightarrow−\mathrm{2}\:=\frac{{d}}{\mathrm{2}}\:\Rightarrow{d}=−\mathrm{4}\:\Rightarrow \\ $$$${F}\left({u}\right)=\frac{\mathrm{1}}{{u}^{\mathrm{2}} }−\frac{\mathrm{4}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$${I}\:=\frac{\mathrm{5}}{\mathrm{2}}\left\{{u}\:+\int\:{F}\left({u}\right){du}\right\} \\ $$$$=\frac{\mathrm{5}}{\mathrm{2}}{u}\:+\frac{\mathrm{5}}{\mathrm{2}}\left(−\frac{\mathrm{1}}{{u}}−\mathrm{4}\:{arctan}\left({u}\right)\right)\:+{C} \\ $$$$=\frac{\mathrm{5}}{\mathrm{2}}{e}^{{t}} −\frac{\mathrm{5}}{\mathrm{2}}{e}^{−{t}} \:\:−\mathrm{10}\:{arctan}\left({e}^{{t}} \right)+{C} \\ $$$${t}={argch}\left(\frac{{x}}{\mathrm{5}}\right)\:={ln}\left(\frac{{x}}{\mathrm{5}}+\sqrt{\frac{{x}^{\mathrm{2}} }{\mathrm{25}}−\mathrm{1}}\right)\:\Rightarrow \\ $$$${I}\:=\frac{\mathrm{5}}{\mathrm{2}}\left\{\:\frac{{x}}{\mathrm{5}}+\sqrt{\frac{{x}^{\mathrm{2}} }{\mathrm{25}}−\mathrm{1}}−\frac{\mathrm{1}}{\frac{{x}}{\mathrm{5}}+\sqrt{\frac{{x}^{\mathrm{2}} }{\mathrm{25}}−\mathrm{1}}}\right\} \\ $$$$−\mathrm{10}\:{arctan}\left(\frac{{x}}{\mathrm{5}}+\sqrt{\frac{{x}^{\mathrm{2}} }{\mathrm{25}}−\mathrm{1}}\right)\:+{C} \\ $$$$ \\ $$
Commented by Kunal12588 last updated on 26/Mar/20
sir in (2) ∫2x^2 dx=(2/3)x^3 ?
$${sir}\:{in}\:\left(\mathrm{2}\right)\:\int\mathrm{2}{x}^{\mathrm{2}} {dx}=\frac{\mathrm{2}}{\mathrm{3}}{x}^{\mathrm{3}} \:? \\ $$
Answered by jagoll last updated on 26/Mar/20
(3) ∫ (dx/( (√(x^2 +4))))? [ x = 2 tan u ] ∫ ((2sec^2 u)/(2sec u)) du = ∫ sec u du = ln ∣sec u + tan u ∣ + c = ln ∣ ((x+(√(x^2 +4)))/2) ∣ + c
$$\left(\mathrm{3}\right)\:\int\:\frac{\mathrm{dx}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{4}}}? \\ $$$$\left[\:\mathrm{x}\:=\:\mathrm{2}\:\mathrm{tan}\:\mathrm{u}\:\right] \\ $$$$\int\:\frac{\mathrm{2sec}\:^{\mathrm{2}} \mathrm{u}}{\mathrm{2sec}\:\mathrm{u}}\:\mathrm{du}\:=\:\int\:\mathrm{sec}\:\mathrm{u}\:\mathrm{du} \\ $$$$=\:\mathrm{ln}\:\mid\mathrm{sec}\:\mathrm{u}\:+\:\mathrm{tan}\:\mathrm{u}\:\mid\:+\:\mathrm{c} \\ $$$$=\:\mathrm{ln}\:\mid\:\frac{\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}\:\mid\:+\:\mathrm{c} \\ $$
Commented by Kunal12588 last updated on 26/Mar/20
=ln ∣x+(√(x^2 +4))∣+C
$$={ln}\:\mid{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\mid+{C} \\ $$
Commented by john santu last updated on 26/Mar/20
same it sir ln ∣((x+(√(x^2 +4)))/2)∣ +c = ln ∣x+(√(x^2 +4 )) ∣ + C
$${same}\:{it}\:{sir} \\ $$$$\mathrm{ln}\:\mid\frac{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}\mid\:+{c}\:=\: \\ $$$$\mathrm{ln}\:\mid{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}\:}\:\mid\:+\:{C} \\ $$
Answered by Kunal12588 last updated on 26/Mar/20
(5)I=∫(dx/( (√(5−4x−2x^2 )))) =(1/( (√2)))∫(dx/( (√(−(x^2 +2x−(5/2)))))) =(1/( (√2)))∫(dx/( (√(−[(x+1)^2 −(7/2)])))) =(1/( (√2)))∫(dx/( (√((7/2)−(x+1)^2 )))) =(1/( (√2)))sin^(−1) (((x+1)/((√7)/( (√2)))))+C =(1/( (√2)))sin^(−1) ((((√2)x+(√2))/( (√7))))+C
$$\left(\mathrm{5}\right){I}=\int\frac{{dx}}{\:\sqrt{\mathrm{5}−\mathrm{4}{x}−\mathrm{2}{x}^{\mathrm{2}} }} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{{dx}}{\:\sqrt{−\left({x}^{\mathrm{2}} +\mathrm{2}{x}−\frac{\mathrm{5}}{\mathrm{2}}\right)}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{{dx}}{\:\sqrt{−\left[\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\frac{\mathrm{7}}{\mathrm{2}}\right]}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{{dx}}{\:\sqrt{\frac{\mathrm{7}}{\mathrm{2}}−\left({x}+\mathrm{1}\right)^{\mathrm{2}} }} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{sin}^{−\mathrm{1}} \left(\frac{{x}+\mathrm{1}}{\frac{\sqrt{\mathrm{7}}}{\:\sqrt{\mathrm{2}}}}\right)+{C} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{sin}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{2}}{x}+\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{7}}}\right)+{C} \\ $$

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