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Question Number 111189 by john santu last updated on 02/Sep/20

(1) ∫ (((x+1)dx)/(x^4 (x−1))) ? (2) (dy/dx) + (y/(x−2)) = 5(x−2)(√y)

$$\:\left(\mathrm{1}\right)\:\:\:\:\:\:\:\int\:\frac{\left({x}+\mathrm{1}\right){dx}}{{x}^{\mathrm{4}} \left({x}−\mathrm{1}\right)}\:?\: \\ $$$$\:\left(\mathrm{2}\right)\:\:\:\:\:\:\frac{{dy}}{{dx}}\:+\:\frac{{y}}{{x}−\mathrm{2}}\:=\:\mathrm{5}\left({x}−\mathrm{2}\right)\sqrt{{y}}\: \\ $$

Answered by Sarah85 last updated on 02/Sep/20

((x+1)/(x^4 (x−1)))=(2/(x−1))−(2/x)−(2/x^2 )−(2/x^3 )−(1/x^4 ) ∫((x+1)/(x^4 (x−1)))dx=ln (((x−1)^2 )/x^2 ) +((6x^2 +3x+1)/(3x^3 ))+C

$$\frac{{x}+\mathrm{1}}{{x}^{\mathrm{4}} \left({x}−\mathrm{1}\right)}=\frac{\mathrm{2}}{{x}−\mathrm{1}}−\frac{\mathrm{2}}{{x}}−\frac{\mathrm{2}}{{x}^{\mathrm{2}} }−\frac{\mathrm{2}}{{x}^{\mathrm{3}} }−\frac{\mathrm{1}}{{x}^{\mathrm{4}} } \\ $$$$\int\frac{{x}+\mathrm{1}}{{x}^{\mathrm{4}} \left({x}−\mathrm{1}\right)}{dx}=\mathrm{ln}\:\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} }\:+\frac{\mathrm{6}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}}{\mathrm{3}{x}^{\mathrm{3}} }+{C} \\ $$

Answered by Dwaipayan Shikari last updated on 02/Sep/20

∫((x−1)/(x^4 (x−1)))+(2/(x^4 (x−1))) −(1/(3x^3 ))−2∫(1/x^3 )((1/x)−(1/(x−1))) =−(1/(3x^3 ))+(2/(3x^3 ))+2∫(1/(x^3 (x−1)))dx =(1/(3x^3 ))−2∫(1/x^2 )((1/x)−(1/(x−1))) =(1/(3x^3 ))+(1/x^2 )+2∫(1/(x^2 (x−1)))dx =(1/(3x^3 ))+(1/x^2 )−2∫(1/x)((1/x)−(1/(x−1))) =(1/(3x^3 ))+(1/x^2 )+(2/x)−2∫(1/x)−(1/(x−1)) =(1/(3x^3 ))+(1/x^2 )+(2/x)−2log(x)+2log(x−1)+C

$$\int\frac{{x}−\mathrm{1}}{{x}^{\mathrm{4}} \left({x}−\mathrm{1}\right)}+\frac{\mathrm{2}}{{x}^{\mathrm{4}} \left({x}−\mathrm{1}\right)} \\ $$$$−\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{3}} }−\mathrm{2}\int\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{x}−\mathrm{1}}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{3}} }+\frac{\mathrm{2}}{\mathrm{3}{x}^{\mathrm{3}} }+\mathrm{2}\int\frac{\mathrm{1}}{{x}^{\mathrm{3}} \left({x}−\mathrm{1}\right)}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{3}} }−\mathrm{2}\int\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{x}−\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{3}} }+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{2}\int\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left({x}−\mathrm{1}\right)}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{3}} }+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{2}\int\frac{\mathrm{1}}{{x}}\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{x}−\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{3}} }+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{2}}{{x}}−\mathrm{2}\int\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{x}−\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{3}} }+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{2}}{{x}}−\mathrm{2}{log}\left({x}\right)+\mathrm{2}{log}\left({x}−\mathrm{1}\right)+{C} \\ $$

Answered by bemath last updated on 02/Sep/20

Answered by ajfour last updated on 02/Sep/20

let y=t^2 , x−2=s ⇒ ((2tdt)/ds)+(t^2 /s)=5st ⇒ if t≠0, (dt/ds)+(t/(2s))=((5s)/2) ⇒ I.F. = e^((1/2)∫(ds/s)) = (√s) t(√s) = ∫((5s(√s))/2) t(√s) = s^2 (√s)+c ⇒ (√y)−(x−2)^2 =(c/( (√(x−2))))

$${let}\:\:{y}={t}^{\mathrm{2}} \:\:\:,\:\:\:{x}−\mathrm{2}={s} \\ $$$$\Rightarrow\:\:\:\frac{\mathrm{2}{tdt}}{{ds}}+\frac{{t}^{\mathrm{2}} }{{s}}=\mathrm{5}{st} \\ $$$$\Rightarrow\:\:\:{if}\:\:{t}\neq\mathrm{0},\:\:\:\:\frac{{dt}}{{ds}}+\frac{{t}}{\mathrm{2}{s}}=\frac{\mathrm{5}{s}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:{I}.{F}.\:=\:{e}^{\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{ds}}{{s}}} =\:\sqrt{{s}} \\ $$$$\:\:{t}\sqrt{{s}}\:=\:\int\frac{\mathrm{5}{s}\sqrt{{s}}}{\mathrm{2}} \\ $$$$\:\:\:{t}\sqrt{{s}}\:=\:{s}^{\mathrm{2}} \sqrt{{s}}+{c} \\ $$$$\Rightarrow\:\:\sqrt{{y}}−\left({x}−\mathrm{2}\right)^{\mathrm{2}} =\frac{{c}}{\:\sqrt{{x}−\mathrm{2}}}\: \\ $$

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