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Question Number 117666 by danielasebhofoh last updated on 13/Oct/20

Commented by bemath last updated on 13/Oct/20

set (√(ax+b)) =t ⇒ ax+b = t^2 x = ((t^2 −b)/a) ⇒dx = ((2t dt)/a) ∫ ((2t dt)/(a(((t^2 −b)/a) + c).t)) = ∫ ((2 dt)/(t^2 −(b−ac))) ∫ ((2 dt)/((t−(√(b−ac)))(t+(√(b−ac))))) now it easy to solve

$$\mathrm{set}\:\sqrt{\mathrm{ax}+\mathrm{b}}\:=\mathrm{t}\:\Rightarrow\:\mathrm{ax}+\mathrm{b}\:=\:\mathrm{t}^{\mathrm{2}} \\ $$$$\mathrm{x}\:=\:\frac{\mathrm{t}^{\mathrm{2}} −\mathrm{b}}{\mathrm{a}}\:\Rightarrow\mathrm{dx}\:=\:\frac{\mathrm{2t}\:\mathrm{dt}}{\mathrm{a}} \\ $$$$\int\:\frac{\mathrm{2t}\:\mathrm{dt}}{\mathrm{a}\left(\frac{\mathrm{t}^{\mathrm{2}} −\mathrm{b}}{\mathrm{a}}\:+\:\mathrm{c}\right).\mathrm{t}}\:=\:\int\:\frac{\mathrm{2}\:\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} −\left(\mathrm{b}−\mathrm{ac}\right)} \\ $$$$\int\:\frac{\mathrm{2}\:\mathrm{dt}}{\left(\mathrm{t}−\sqrt{\mathrm{b}−\mathrm{ac}}\right)\left(\mathrm{t}+\sqrt{\mathrm{b}−\mathrm{ac}}\right)}\: \\ $$$$\mathrm{now}\:\mathrm{it}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve} \\ $$

Commented by MJS_new last updated on 13/Oct/20

typo: ∫((2dt)/(t^2 −(b−ac)))

$$\mathrm{typo}:\:\int\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} −\left({b}−{ac}\right)} \\ $$

Commented by bemath last updated on 13/Oct/20

haha..yes sir thank you

$$\mathrm{haha}..\mathrm{yes}\:\mathrm{sir}\:\mathrm{thank}\:\mathrm{you} \\ $$

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