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Question Number 36444 by rahul 19 last updated on 02/Jun/18

∫ (dx/((2x−7)(√((x−3)(x−4))))) = ?

$$\int\:\frac{\mathrm{d}{x}}{\left(\mathrm{2}{x}−\mathrm{7}\right)\sqrt{\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right)}}\:=\:? \\ $$

Commented by math1967 last updated on 02/Jun/18

let2x−7=(1/z) ∴2dx=−(dz/z^2 ) −∫(dz/(2z^2 ×(1/z)(√((((1+z))/(2z))×(((1−z))/(2z)))))) −∫(dz/(√(1−z^2 ))) −sin^(−1) z+c=−sin^(−1) (1/(2x−7))+c

$${let}\mathrm{2}{x}−\mathrm{7}=\frac{\mathrm{1}}{{z}}\:\:\therefore\mathrm{2}{dx}=−\frac{{dz}}{{z}^{\mathrm{2}} } \\ $$$$−\int\frac{{dz}}{\mathrm{2}{z}^{\mathrm{2}} ×\frac{\mathrm{1}}{{z}}\sqrt{\frac{\left(\mathrm{1}+{z}\right)}{\mathrm{2}{z}}×\frac{\left(\mathrm{1}−{z}\right)}{\mathrm{2}{z}}}} \\ $$$$−\int\frac{{dz}}{\sqrt{\mathrm{1}−{z}^{\mathrm{2}} }} \\ $$$$−\mathrm{sin}^{−\mathrm{1}} {z}+{c}=−\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}{x}−\mathrm{7}}+{c} \\ $$

Commented by math1967 last updated on 02/Jun/18

Am I correct?

$${Am}\:{I}\:{correct}? \\ $$

Commented by MJS last updated on 02/Jun/18

indeed correct!

$$\mathrm{indeed}\:\mathrm{correct}! \\ $$

Commented by rahul 19 last updated on 02/Jun/18

Thanks for confirming!

$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{confirming}! \\ $$

Commented by MJS last updated on 02/Jun/18

another way: ∫(dx/((2x−7)(√(x^2 −7x+12))))= [t=(√(x^2 −7x+12)) → dx=((2(√(x^2 −7x+12)))/(2x−7))dt] =2∫(dt/(4t^2 +1))= [u=2t → dt=(du/2)] =∫(du/(u^2 +1))=arctan u=arctan 2t= =arctan 2(√((x−3)(x−4)))+C

$$\mathrm{another}\:\mathrm{way}: \\ $$$$\int\frac{{dx}}{\left(\mathrm{2}{x}−\mathrm{7}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{12}}}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{12}}\:\rightarrow\:{dx}=\frac{\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{12}}}{\mathrm{2}{x}−\mathrm{7}}{dt}\right] \\ $$$$=\mathrm{2}\int\frac{{dt}}{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$\:\:\:\:\:\left[{u}=\mathrm{2}{t}\:\rightarrow\:{dt}=\frac{{du}}{\mathrm{2}}\right] \\ $$$$=\int\frac{{du}}{{u}^{\mathrm{2}} +\mathrm{1}}=\mathrm{arctan}\:{u}=\mathrm{arctan}\:\mathrm{2}{t}= \\ $$$$=\mathrm{arctan}\:\mathrm{2}\sqrt{\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right)}+{C} \\ $$

Commented by math1967 last updated on 02/Jun/18

It is more easy.Thanks sir

$${It}\:{is}\:{more}\:{easy}.{Thanks}\:{sir} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 02/Jun/18

(x−3)(x−4)=x^2 −7x+12 not x^2 +7x+12

$$\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right)={x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{12}\:\:\:{not}\:{x}^{\mathrm{2}} +\mathrm{7}{x}+\mathrm{12} \\ $$

Commented by math1967 last updated on 02/Jun/18

you are correct Tanmay sir

$${you}\:{are}\:{correct}\:{Tanmay}\:{sir} \\ $$

Commented by MJS last updated on 02/Jun/18

sorry, typo. I corrected it

$$\mathrm{sorry},\:\mathrm{typo}.\:\mathrm{I}\:\mathrm{corrected}\:\mathrm{it} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 02/Jun/18

Commented by prof Abdo imad last updated on 03/Jun/18

changement 2x−7=t give I = ∫ (1/(t(√((((t+7)/2)−3)(((t+7)/2) −4)))))(dt/2) = (1/2)∫ ((2dt)/(t(√((t+1)(t−1))))) = ∫ (dt/(t(√(t^2 −1)))) and changement t =ch(u) give I = ∫ ((sh(u))/(ch(u)shu))du = ∫ (du/(chu)) =∫ ((2du)/(e^u +e^(−u) )) =_(e^u =x) ∫ (2/(x +(1/x))) (dx/x) = 2 ∫ (dx/(x^2 +1)) = 2arctanx +c =2arctan( e^u ) = 2arctan ( e^(argch(t)) ) +c = 2arctan (ln(t +(√(t^2 −1))) +c =2arctan{ln(2x−7 +(√((2x−7)^2 −1)))} +c .

$${changement}\:\:\mathrm{2}{x}−\mathrm{7}={t}\:{give} \\ $$$${I}\:=\:\int\:\:\:\:\:\:\frac{\mathrm{1}}{{t}\sqrt{\left(\frac{{t}+\mathrm{7}}{\mathrm{2}}−\mathrm{3}\right)\left(\frac{{t}+\mathrm{7}}{\mathrm{2}}\:−\mathrm{4}\right)}}\frac{{dt}}{\mathrm{2}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\:\frac{\mathrm{2}{dt}}{{t}\sqrt{\left({t}+\mathrm{1}\right)\left({t}−\mathrm{1}\right)}} \\ $$$$=\:\int\:\:\:\:\:\frac{{dt}}{{t}\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}}\:{and}\:{changement}\:{t}\:={ch}\left({u}\right)\:{give} \\ $$$${I}\:=\:\int\:\:\:\:\:\frac{{sh}\left({u}\right)}{{ch}\left({u}\right){shu}}{du}\:=\:\int\:\:\:\:\:\frac{{du}}{{chu}} \\ $$$$=\int\:\:\:\:\frac{\mathrm{2}{du}}{{e}^{{u}} \:+{e}^{−{u}} }\:=_{{e}^{{u}} ={x}} \:\:\int\:\:\:\:\:\frac{\mathrm{2}}{{x}\:+\frac{\mathrm{1}}{{x}}}\:\frac{{dx}}{{x}} \\ $$$$=\:\mathrm{2}\:\int\:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{1}}\:=\:\mathrm{2}{arctanx}\:+{c} \\ $$$$=\mathrm{2}{arctan}\left(\:{e}^{{u}} \right)\:=\:\mathrm{2}{arctan}\:\left(\:{e}^{{argch}\left({t}\right)} \right)\:+{c} \\ $$$$=\:\mathrm{2}{arctan}\:\left({ln}\left({t}\:+\sqrt{\left.{t}^{\mathrm{2}} −\mathrm{1}\right)}\:\:+{c}\right.\right. \\ $$$$=\mathrm{2}{arctan}\left\{{ln}\left(\mathrm{2}{x}−\mathrm{7}\:+\sqrt{\left.\left(\mathrm{2}{x}−\mathrm{7}\right)^{\mathrm{2}} \:−\mathrm{1}\right)}\right\}\:+{c}\:.\right. \\ $$$$ \\ $$

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