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Question Number 57749 by maxmathsup by imad last updated on 11/Apr/19
find ∫ (dx/(x^2 (√(9+x^2 ))))
$${find}\:\int\:\:\frac{{dx}}{{x}^{\mathrm{2}} \sqrt{\mathrm{9}+{x}^{\mathrm{2}} }} \\ $$
Commented by maxmathsup by imad last updated on 11/Apr/19
let A =∫ (dx/(x^2 (√(9+x^2 )))) changement x =3sh(t) give A =∫ ((3ch(t))/(9sh^2 t3ch(t))) dt =(1/9) ∫ (dt/(sh^2 t)) ⇒ 9A =∫ ((2dt)/(ch(2t)−1)) =∫ ((2dt)/(((e^(2t) +e^(−2t) )/2)−1)) =4 ∫ (dt/(e^(2t) +e^(−2t) −2)) =_(e^(2t) =u) 4 ∫ (du/(2u(u +u^(−1) −2))) ( t =((ln(u))/2)) =2 ∫ (du/(u^2 +1−2u)) =2 ∫ (du/((u−1)^2 )) =((−2)/(u−1)) +c = ((−2)/(e^(2t) −1)) +c we have sh(t) =(x/3) ⇒t =argsh((x/3)) =ln((x/3)+(√(1+(x^2 /9)))) ⇒ e^(2t) =((x/3) +(√(1+(x^2 /9))))^2 =(1/9)(x+(√(x^2 +9)))^2 ⇒ 9A =(2/(1−(1/9)(x+(√(x^2 +9)))^2 )) +c = ((18)/(9−(x+(√(x^2 +9)))^2 )) +c ⇒ A =(2/(9−(x+(√(x^2 +9)))^2 )) +C
$${let}\:{A}\:=\int\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \sqrt{\mathrm{9}+{x}^{\mathrm{2}} }}\:\:\:{changement}\:{x}\:=\mathrm{3}{sh}\left({t}\right)\:{give}\: \\ $$$${A}\:=\int\:\:\:\frac{\mathrm{3}{ch}\left({t}\right)}{\mathrm{9}{sh}^{\mathrm{2}} {t}\mathrm{3}{ch}\left({t}\right)}\:{dt}\:=\frac{\mathrm{1}}{\mathrm{9}}\:\int\:\:\frac{{dt}}{{sh}^{\mathrm{2}} {t}}\:\Rightarrow\:\mathrm{9}{A}\:=\int\:\:\:\frac{\mathrm{2}{dt}}{{ch}\left(\mathrm{2}{t}\right)−\mathrm{1}} \\ $$$$=\int\:\:\:\frac{\mathrm{2}{dt}}{\frac{{e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} }{\mathrm{2}}−\mathrm{1}}\:=\mathrm{4}\:\int\:\:\:\frac{{dt}}{{e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} \:−\mathrm{2}}\:=_{{e}^{\mathrm{2}{t}} \:={u}} \:\:\:\mathrm{4}\:\int\:\:\frac{{du}}{\mathrm{2}{u}\left({u}\:+{u}^{−\mathrm{1}} −\mathrm{2}\right)}\:\:\:\left(\:{t}\:=\frac{{ln}\left({u}\right)}{\mathrm{2}}\right) \\ $$$$=\mathrm{2}\:\int\:\:\:\frac{{du}}{{u}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{2}{u}}\:=\mathrm{2}\:\int\:\:\:\frac{{du}}{\left({u}−\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{−\mathrm{2}}{{u}−\mathrm{1}}\:+{c}\:=\:\frac{−\mathrm{2}}{{e}^{\mathrm{2}{t}} −\mathrm{1}}\:+{c} \\ $$$${we}\:{have}\:{sh}\left({t}\right)\:=\frac{{x}}{\mathrm{3}}\:\Rightarrow{t}\:={argsh}\left(\frac{{x}}{\mathrm{3}}\right)\:={ln}\left(\frac{{x}}{\mathrm{3}}+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{9}}}\right)\:\Rightarrow \\ $$$${e}^{\mathrm{2}{t}} \:=\left(\frac{{x}}{\mathrm{3}}\:+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{9}}}\right)^{\mathrm{2}} \:=\frac{\mathrm{1}}{\mathrm{9}}\left({x}+\sqrt{{x}^{\mathrm{2}} \:+\mathrm{9}}\right)^{\mathrm{2}} \:\Rightarrow \\ $$$$\mathrm{9}{A}\:=\frac{\mathrm{2}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{9}}\left({x}+\sqrt{{x}^{\mathrm{2}} \:+\mathrm{9}}\right)^{\mathrm{2}} }\:+{c}\:=\:\frac{\mathrm{18}}{\mathrm{9}−\left({x}+\sqrt{{x}^{\mathrm{2}} \:+\mathrm{9}}\right)^{\mathrm{2}} }\:+{c}\:\:\Rightarrow \\ $$$${A}\:=\frac{\mathrm{2}}{\mathrm{9}−\left({x}+\sqrt{{x}^{\mathrm{2}} \:+\mathrm{9}}\right)^{\mathrm{2}} }\:+{C} \\ $$
Commented by MJS last updated on 11/Apr/19
(2/(9−(x+(√(x^2 +9)))^2 ))=(2/(−2x^2 −2x(√(x^2 +9))))= =−(1/(x(x+(√(x^2 +9)))))=−((x−(√(x^2 +9)))/(x(x+(√(x^2 +9)))(x−(√(x^2 +9)))))= =−((x−(√(x^2 +9)))/(−9x))=(1/9)−((√(x^2 +9))/(9x)) A=−((√(x^2 +9))/(9x))+C
$$\frac{\mathrm{2}}{\mathrm{9}−\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{9}}\right)^{\mathrm{2}} }=\frac{\mathrm{2}}{−\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{9}}}= \\ $$$$=−\frac{\mathrm{1}}{{x}\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{9}}\right)}=−\frac{{x}−\sqrt{{x}^{\mathrm{2}} +\mathrm{9}}}{{x}\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{9}}\right)\left({x}−\sqrt{{x}^{\mathrm{2}} +\mathrm{9}}\right)}= \\ $$$$=−\frac{{x}−\sqrt{{x}^{\mathrm{2}} +\mathrm{9}}}{−\mathrm{9}{x}}=\frac{\mathrm{1}}{\mathrm{9}}−\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{9}}}{\mathrm{9}{x}} \\ $$$${A}=−\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{9}}}{\mathrm{9}{x}}+{C} \\ $$
Commented by maxmathsup by imad last updated on 11/Apr/19
thanks sir for completing calculus.
$${thanks}\:{sir}\:{for}\:{completing}\:{calculus}. \\ $$
Answered by MJS last updated on 11/Apr/19
(d/dx)[(√(x^2 +9))]=(x/( (√(x^2 +9)))) (d/dx)[((√(x^2 +9))/(f(x)))]=((xf(x)−(x^2 +9)f′(x))/( (√(x^2 +9))(f(x))^2 )) trying f(x)=x this leads to −(9/(x^2 (√(x^2 +9)))) trying f(x)=−9x this leads to (1/(x^2 (√(x^2 +9)))) ⇒ ∫(dx/(x^2 (√(x^2 +9))))=−((√(x^2 +9))/(9x))+C
$$\frac{{d}}{{dx}}\left[\sqrt{{x}^{\mathrm{2}} +\mathrm{9}}\right]=\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{9}}} \\ $$$$\frac{{d}}{{dx}}\left[\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{9}}}{{f}\left({x}\right)}\right]=\frac{{xf}\left({x}\right)−\left({x}^{\mathrm{2}} +\mathrm{9}\right){f}'\left({x}\right)}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{9}}\left({f}\left({x}\right)\right)^{\mathrm{2}} } \\ $$$$\mathrm{trying}\:{f}\left({x}\right)={x}\:\mathrm{this}\:\mathrm{leads}\:\mathrm{to}\:−\frac{\mathrm{9}}{{x}^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} +\mathrm{9}}} \\ $$$$\mathrm{trying}\:{f}\left({x}\right)=−\mathrm{9}{x}\:\mathrm{this}\:\mathrm{leads}\:\mathrm{to}\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} +\mathrm{9}}} \\ $$$$\Rightarrow\:\int\frac{{dx}}{{x}^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} +\mathrm{9}}}=−\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{9}}}{\mathrm{9}{x}}+{C} \\ $$

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