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7x-6-x-2-25-x-3-2-4-dx-

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Question Number 35909 by ajfour last updated on 25/May/18
∫((7x−6)/((x^2 +25)(√((x−3)^2 +4)))) dx = ?
$$\int\frac{\mathrm{7}{x}−\mathrm{6}}{\left({x}^{\mathrm{2}} +\mathrm{25}\right)\sqrt{\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{4}}}\:{dx}\:=\:? \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 26/May/18
this nut is hard to crack...i am fighting..hope reach the destination to get it solved...
$${this}\:{nut}\:{is}\:{hard}\:{to}\:{crack}…{i}\:{am}\:{fighting}..{hope} \\ $$$${reach}\:{the}\:{destination}\:{to}\:{get}\:{it}\:{solved}… \\ $$
Commented by rahul 19 last updated on 26/May/18
Yes sir ! If there is no ( 7x−6 ) in numerator then it is cakewalk!
$${Yes}\:{sir}\:! \\ $$$${If}\:{there}\:{is}\:{no}\:\left(\:\mathrm{7}{x}−\mathrm{6}\:\right)\:{in}\:{numerator} \\ $$$${then}\:{it}\:{is}\:{cakewalk}! \\ $$
Commented by prof Abdo imad last updated on 26/May/18
tbere is no cake ni chocolat in maths...
$${tbere}\:{is}\:{no}\:{cake}\:{ni}\:{chocolat}\:{in}\:{maths}… \\ $$
Commented by abdo mathsup 649 cc last updated on 27/May/18
changement x−3=2sht give I = ∫ ((7(3+2sht)−6)/({ (2sht+3)^2 +25}2 cht)) 2chtdt I = ∫ ((14sht +15)/({4sh^2 t +12sht +34)))dt = ∫ ((14((e^t −e^(−t) )/2) +15)/(4 (((e^t −e^(−t) )/2))^2 +12 ((e^t −e^(−t) )/2) +34))dt = ∫ ((7e^t −7 e^(−t) +15)/(e^(2t) +e^(−2t) −2 +6 e^t −6 e^(−t) +34))dt =∫ ((7e^t −7 e^(−t) +15)/(e^(2t) +e^(−2t) +6 e^t −6 e^(−t) +32))dt changement e^t =x give I = ∫ ((7x−(7/x) +15)/(x^2 +(1/x^2 ) +6x −(6/x) +32)) (dx/x) = ∫ ((7x^2 +15x −7)/(x^4 +1 +6x^3 −6x +32x^2 ))dx = ∫ ((7x^2 +15x −7)/(x^4 +6x^3 +32x^2 −6x +1))dx the roots of polynom p(x)= x^4 +6x^3 +32 x^2 −6x +1 are the complex z_1 ∼ −3,093+4,853 i z_2 ∼−3,093 −4,853 i =z_1 ^− z_3 ∼ 0,093 +0,146 i z_4 ∼0,093 −0,146 i= z_3 ^− p(x) ∼ (z−_ z_1 )(x−z_1 ^− )(x−z_2 )(x −z_2 ^− ) ∼ (x^2 +2.3,093x+(√((3,093)^2 +(4,853)^2 ))). ( x^2 +2.0,093 x +(√((0,093)^2 +(0,146)^2 ))) the value of this integral is obtained after decomposing the fraction F(x) = ((7x^2 +15x −7)/(p(x))) at form F(x) = ((ax+b)/(x^2 −2Re(z_1 )x +∣z_1 ∣^2 )) + ((cx+d)/(x^2 −2Re(z_2 )x +∣z_2 ∣^( )) ....
$${changement}\:{x}−\mathrm{3}=\mathrm{2}{sht}\:{give}\: \\ $$$${I}\:=\:\int\:\:\frac{\mathrm{7}\left(\mathrm{3}+\mathrm{2}{sht}\right)−\mathrm{6}}{\left\{\:\left(\mathrm{2}{sht}+\mathrm{3}\right)^{\mathrm{2}} \:+\mathrm{25}\right\}\mathrm{2}\:{cht}}\:\mathrm{2}{chtdt} \\ $$$${I}\:=\:\int\:\:\:\:\frac{\mathrm{14}{sht}\:\:+\mathrm{15}}{\left\{\mathrm{4}{sh}^{\mathrm{2}} {t}\:+\mathrm{12}{sht}\:\:+\mathrm{34}\right)}{dt} \\ $$$$=\:\int\:\:\:\frac{\mathrm{14}\frac{{e}^{{t}} −{e}^{−{t}} }{\mathrm{2}}\:+\mathrm{15}}{\mathrm{4}\:\left(\frac{{e}^{{t}} \:−{e}^{−{t}} }{\mathrm{2}}\right)^{\mathrm{2}} \:+\mathrm{12}\:\frac{{e}^{{t}} \:−{e}^{−{t}} }{\mathrm{2}}\:+\mathrm{34}}{dt} \\ $$$$=\:\int\:\:\:\frac{\mathrm{7}{e}^{{t}} \:−\mathrm{7}\:{e}^{−{t}} \:+\mathrm{15}}{{e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} −\mathrm{2}\:+\mathrm{6}\:{e}^{{t}} \:−\mathrm{6}\:{e}^{−{t}} \:+\mathrm{34}}{dt} \\ $$$$=\int\:\:\:\frac{\mathrm{7}{e}^{{t}} \:−\mathrm{7}\:{e}^{−{t}} \:+\mathrm{15}}{{e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} \:\:+\mathrm{6}\:{e}^{{t}} \:−\mathrm{6}\:{e}^{−{t}} \:\:+\mathrm{32}}{dt}\:{changement} \\ $$$${e}^{{t}} \:={x}\:{give} \\ $$$${I}\:=\:\int\:\:\frac{\mathrm{7}{x}−\frac{\mathrm{7}}{{x}}\:+\mathrm{15}}{{x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\mathrm{6}{x}\:−\frac{\mathrm{6}}{{x}}\:+\mathrm{32}}\:\frac{{dx}}{{x}} \\ $$$$=\:\int\:\:\:\:\frac{\mathrm{7}{x}^{\mathrm{2}} \:+\mathrm{15}{x}\:−\mathrm{7}}{{x}^{\mathrm{4}} \:\:\:+\mathrm{1}\:+\mathrm{6}{x}^{\mathrm{3}} \:−\mathrm{6}{x}\:+\mathrm{32}{x}^{\mathrm{2}} }{dx} \\ $$$$=\:\int\:\:\:\frac{\mathrm{7}{x}^{\mathrm{2}} \:+\mathrm{15}{x}\:−\mathrm{7}}{{x}^{\mathrm{4}} \:+\mathrm{6}{x}^{\mathrm{3}} \:+\mathrm{32}{x}^{\mathrm{2}} \:−\mathrm{6}{x}\:+\mathrm{1}}{dx}\:{the}\:{roots}\:{of} \\ $$$${polynom}\:{p}\left({x}\right)=\:{x}^{\mathrm{4}} \:+\mathrm{6}{x}^{\mathrm{3}} \:+\mathrm{32}\:{x}^{\mathrm{2}} \:−\mathrm{6}{x}\:+\mathrm{1}\:{are} \\ $$$${the}\:{complex}\:\:{z}_{\mathrm{1}} \:\sim\:−\mathrm{3},\mathrm{093}+\mathrm{4},\mathrm{853}\:{i} \\ $$$${z}_{\mathrm{2}} \sim−\mathrm{3},\mathrm{093}\:−\mathrm{4},\mathrm{853}\:{i}\:=\overset{−} {{z}}_{\mathrm{1}} \\ $$$${z}_{\mathrm{3}} \:\sim\:\mathrm{0},\mathrm{093}\:+\mathrm{0},\mathrm{146}\:{i} \\ $$$${z}_{\mathrm{4}} \:\sim\mathrm{0},\mathrm{093}\:−\mathrm{0},\mathrm{146}\:{i}=\:\overset{−} {{z}}_{\mathrm{3}} \\ $$$${p}\left({x}\right)\:\sim\:\left({z}−_{} {z}_{\mathrm{1}} \right)\left({x}−\overset{−} {{z}}_{\mathrm{1}} \right)\left({x}−{z}_{\mathrm{2}} \right)\left({x}\:−\overset{−} {{z}}_{\mathrm{2}} \right) \\ $$$$\sim\:\left({x}^{\mathrm{2}} \:\:+\mathrm{2}.\mathrm{3},\mathrm{093}{x}+\sqrt{\left(\mathrm{3},\mathrm{093}\right)^{\mathrm{2}} \:+\left(\mathrm{4},\mathrm{853}\right)^{\mathrm{2}} }\right). \\ $$$$\left(\:{x}^{\mathrm{2}} \:\:\:+\mathrm{2}.\mathrm{0},\mathrm{093}\:{x}\:+\sqrt{\left(\mathrm{0},\mathrm{093}\right)^{\mathrm{2}} \:+\left(\mathrm{0},\mathrm{146}\right)^{\mathrm{2}} }\right) \\ $$$$\:{the}\:{value}\:{of}\:{this}\:{integral}\:{is}\:{obtained}\:\:{after} \\ $$$${decomposing}\:{the}\:{fraction} \\ $$$${F}\left({x}\right)\:=\:\frac{\mathrm{7}{x}^{\mathrm{2}} \:+\mathrm{15}{x}\:−\mathrm{7}}{{p}\left({x}\right)}\:{at}\:{form} \\ $$$${F}\left({x}\right)\:=\:\frac{{ax}+{b}}{{x}^{\mathrm{2}} \:−\mathrm{2}{Re}\left({z}_{\mathrm{1}} \right){x}\:+\mid{z}_{\mathrm{1}} \mid^{\mathrm{2}} }\:+\:\frac{{cx}+{d}}{{x}^{\mathrm{2}} \:−\mathrm{2}{Re}\left({z}_{\mathrm{2}} \right){x}\:+\mid{z}_{\mathrm{2}} \mid^{\left(\right.} } \\ $$$$…. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 26/May/18
∫((7x−6)/((x^2 +25)(√(x^2 −6x+13))))dx let t^2 =((x^2 −6x+13)/(x^2 +25)) 2lnt=ln(x^2 −6x+13)−ln(x^2 +25) (2/t)dt={((2x−6)/((x^2 −6x+13)))−((2x)/(x^2 +25))}dx (2/t)dt=((2x^3 +50x−6x^2 −150−2x^3 +12x^2 −26x)/((x^2 +25)(x^2 −6x+13)))dx (2/t)dt=((6x^2 +24x−150)/((x^2 +25)(x^2 −6x+13)))dx dx=(2/t)×(((x^2 +25)(x^2 −6x+13))/(6(x^2 +4x−25)))dt ∫((7x−6)/((x^2 +25)(√(x^2 −6x+13))))(2/t)×(((x^2 +25)(x^2 −6x+13)/(6(x^2 +4x−25))) contd puting value of t ∫((7x−6)/((x^2 +25)(√(x^2 −6x+13))))×((2(√(x^2 +25)))/( (√(x^2 −6x+13))))× (((x^2 +25)(x^2 −6x+13))/(6(x^2 +4x−25)))dx (1/3)∫(((7x−6)(√(x^2 +25)))/(x^2 +4x−25))dx contd
$$ \\ $$$$ \\ $$$$\int\frac{\mathrm{7}{x}−\mathrm{6}}{\left({x}^{\mathrm{2}} +\mathrm{25}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{13}}}{dx} \\ $$$${let}\:{t}^{\mathrm{2}} =\frac{{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{13}}{{x}^{\mathrm{2}} +\mathrm{25}} \\ $$$$\mathrm{2}{lnt}={ln}\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{13}\right)−{ln}\left({x}^{\mathrm{2}} +\mathrm{25}\right) \\ $$$$\frac{\mathrm{2}}{{t}}{dt}=\left\{\frac{\mathrm{2}{x}−\mathrm{6}}{\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{13}\right)}−\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{25}}\right\}{dx}\: \\ $$$$\:\frac{\mathrm{2}}{{t}}{dt}=\frac{\mathrm{2}{x}^{\mathrm{3}} +\mathrm{50}{x}−\mathrm{6}{x}^{\mathrm{2}} −\mathrm{150}−\mathrm{2}{x}^{\mathrm{3}} +\mathrm{12}{x}^{\mathrm{2}} −\mathrm{26}{x}}{\left({x}^{\mathrm{2}} +\mathrm{25}\right)\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{13}\right)}{dx}\:\:\:\:\: \\ $$$$\frac{\mathrm{2}}{{t}}{dt}=\frac{\mathrm{6}{x}^{\mathrm{2}} +\mathrm{24}{x}−\mathrm{150}}{\left({x}^{\mathrm{2}} +\mathrm{25}\right)\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{13}\right)}{dx} \\ $$$${dx}=\frac{\mathrm{2}}{{t}}×\frac{\left({x}^{\mathrm{2}} +\mathrm{25}\right)\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{13}\right)}{\mathrm{6}\left({x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{25}\right)}{dt} \\ $$$$\int\frac{\mathrm{7}{x}−\mathrm{6}}{\left({x}^{\mathrm{2}} +\mathrm{25}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{13}}}\frac{\mathrm{2}}{{t}}×\frac{\left({x}^{\mathrm{2}} +\mathrm{25}\right)\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{13}\right.}{\mathrm{6}\left({x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{25}\right)} \\ $$$${contd}\:{puting}\:{value}\:{of}\:{t} \\ $$$$\int\frac{\mathrm{7}{x}−\mathrm{6}}{\left({x}^{\mathrm{2}} +\mathrm{25}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{13}}}×\frac{\mathrm{2}\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{13}}}× \\ $$$$\frac{\left({x}^{\mathrm{2}} +\mathrm{25}\right)\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{13}\right)}{\mathrm{6}\left({x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{25}\right)}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\left(\mathrm{7}{x}−\mathrm{6}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}}{{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{25}}{dx} \\ $$$${contd} \\ $$
Commented by ajfour last updated on 26/May/18
thanks for the attempt sir.
$${thanks}\:{for}\:{the}\:{attempt}\:{sir}. \\ $$

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