Question Number 169677 by cortano1 last updated on 06/May/22
$$\:\:\:\:{M}\:=\:\int\:\frac{{dx}}{\left({x}−\mathrm{4}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{8}}}\:=? \\ $$
Answered by MJS_new last updated on 06/May/22
![∫(dx/((x+c)(√(x^2 +ax+b))))= [t=(√(x^2 +ax+b))+x+(a/2) → dx=((√(x^2 +ax+b))/t)dt] =2∫(dt/(t^2 +(2c−a)t+(a^2 /4)−b)) here a=−6∧b=8∧c=−4 ⇒ M=2∫(dt/((t^2 −1)^2 ))=−(2/(t−1))=−(2/(x−4+(√(x^2 −6x+8))))+C](https://www.tinkutara.com/question/Q169689.png)
$$\int\frac{{dx}}{\left({x}+{c}\right)\sqrt{{x}^{\mathrm{2}} +{ax}+{b}}}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{{x}^{\mathrm{2}} +{ax}+{b}}+{x}+\frac{{a}}{\mathrm{2}}\:\rightarrow\:{dx}=\frac{\sqrt{{x}^{\mathrm{2}} +{ax}+{b}}}{{t}}{dt}\right] \\ $$$$=\mathrm{2}\int\frac{{dt}}{{t}^{\mathrm{2}} +\left(\mathrm{2}{c}−{a}\right){t}+\frac{{a}^{\mathrm{2}} }{\mathrm{4}}−{b}} \\ $$$$\mathrm{here}\:{a}=−\mathrm{6}\wedge{b}=\mathrm{8}\wedge{c}=−\mathrm{4} \\ $$$$\Rightarrow \\ $$$${M}=\mathrm{2}\int\frac{{dt}}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }=−\frac{\mathrm{2}}{{t}−\mathrm{1}}=−\frac{\mathrm{2}}{{x}−\mathrm{4}+\sqrt{{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{8}}}+{C} \\ $$
Answered by cortano1 last updated on 06/May/22
