Menu Close

let-f-x-1-x-1-2-x-1-1-find-f-1-x-2-find-f-x-dx-3-find-f-1-x-dx-4-find-dx-f-1-x-

Tinku Tara 0 Comments
Pin




Question Number 43809 by maxmathsup by imad last updated on 15/Sep/18
let f(x) =((1−(√(x−1)))/(2+(√(x−1)))) 1) find f^(−1) (x) 2) find ∫ f(x)dx 3) find ∫ f^(−1) (x)dx 4) find ∫ (dx/(f^(−1) (x))) .
$${let}\:{f}\left({x}\right)\:=\frac{\mathrm{1}−\sqrt{{x}−\mathrm{1}}}{\mathrm{2}+\sqrt{{x}−\mathrm{1}}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{f}^{−\mathrm{1}} \left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:\int\:{f}\left({x}\right){dx} \\ $$$$\left.\mathrm{3}\right)\:{find}\:\:\int\:\:{f}^{−\mathrm{1}} \left({x}\right){dx} \\ $$$$\left.\mathrm{4}\right)\:{find}\:\int\:\:\:\frac{{dx}}{{f}^{−\mathrm{1}} \left({x}\right)}\:. \\ $$
Commented by maxmathsup by imad last updated on 17/Sep/18
1) f is defined on [1,+∞[let f(x)=y ⇒x=f^(−1) (y) f(x)=y ⇒((1−(√(x−1)))/(2+(√(x−1))))=y ⇒1−(√(x−1))=2y +y(√(x−1)) ⇒1−2y =(1+y)(√(x−1))⇒ (√(x−1))=((1−2y)/(1+y)) ⇒x−1=(((1−2y)/(1+y)))^2 ⇒x =(((2y−1)/(y+1)))^2 +1 ⇒f^(−1) (x)=(((2x−1)/(x+1)))^2 +1 2) changement (√(x−1))=t give x−1=t^2 ⇒ ∫ f(x)dx = ∫ ((1−t)/(2+t))(2t)dt = 2 ∫ ((t^2 −t)/(t+2))dt =2 ∫ ((t^2 +2t −3t)/(t+2))dt =2 ∫ ((t(t+2))/(t+2))dt −6 ∫ ((t+2−2)/(t+2))dt = t^2 −6t +12 ln∣t+2∣ +c ⇒ ∫ f(x)dx =x−1 −6(√(x−1)) +12ln∣(√(x−1)) +2∣ +c .
$$\left.\mathrm{1}\right)\:{f}\:{is}\:{defined}\:{on}\:\left[\mathrm{1},+\infty\left[{let}\:{f}\left({x}\right)={y}\:\Rightarrow{x}={f}^{−\mathrm{1}} \left({y}\right)\right.\right. \\ $$$${f}\left({x}\right)={y}\:\Rightarrow\frac{\mathrm{1}−\sqrt{{x}−\mathrm{1}}}{\mathrm{2}+\sqrt{{x}−\mathrm{1}}}={y}\:\Rightarrow\mathrm{1}−\sqrt{{x}−\mathrm{1}}=\mathrm{2}{y}\:+{y}\sqrt{{x}−\mathrm{1}}\:\Rightarrow\mathrm{1}−\mathrm{2}{y}\:=\left(\mathrm{1}+{y}\right)\sqrt{{x}−\mathrm{1}}\Rightarrow \\ $$$$\sqrt{{x}−\mathrm{1}}=\frac{\mathrm{1}−\mathrm{2}{y}}{\mathrm{1}+{y}}\:\Rightarrow{x}−\mathrm{1}=\left(\frac{\mathrm{1}−\mathrm{2}{y}}{\mathrm{1}+{y}}\right)^{\mathrm{2}} \:\Rightarrow{x}\:=\left(\frac{\mathrm{2}{y}−\mathrm{1}}{{y}+\mathrm{1}}\right)^{\mathrm{2}} \:+\mathrm{1}\:\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=\left(\frac{\mathrm{2}{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)^{\mathrm{2}} \:+\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:{changement}\:\sqrt{{x}−\mathrm{1}}={t}\:{give}\:{x}−\mathrm{1}={t}^{\mathrm{2}} \:\Rightarrow \\ $$$$\int\:{f}\left({x}\right){dx}\:=\:\int\:\:\frac{\mathrm{1}−{t}}{\mathrm{2}+{t}}\left(\mathrm{2}{t}\right){dt}\:=\:\mathrm{2}\:\int\:\:\:\:\frac{{t}^{\mathrm{2}} \:−{t}}{{t}+\mathrm{2}}{dt}\:=\mathrm{2}\:\int\:\frac{{t}^{\mathrm{2}} \:+\mathrm{2}{t}\:−\mathrm{3}{t}}{{t}+\mathrm{2}}{dt} \\ $$$$=\mathrm{2}\:\int\:\frac{{t}\left({t}+\mathrm{2}\right)}{{t}+\mathrm{2}}{dt}\:\:−\mathrm{6}\:\int\:\:\frac{{t}+\mathrm{2}−\mathrm{2}}{{t}+\mathrm{2}}{dt}\:=\:{t}^{\mathrm{2}} \:−\mathrm{6}{t}\:\:+\mathrm{12}\:{ln}\mid{t}+\mathrm{2}\mid\:+{c}\:\Rightarrow \\ $$$$\int\:{f}\left({x}\right){dx}\:={x}−\mathrm{1}\:−\mathrm{6}\sqrt{{x}−\mathrm{1}}\:+\mathrm{12}{ln}\mid\sqrt{{x}−\mathrm{1}}\:+\mathrm{2}\mid\:+{c}\:. \\ $$$$ \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 17/Sep/18
3) ∫ f^(−1) (x)dx = ∫ ( ((4x^2 −4x +1)/(x^2 +2x+1)) +1)dx= ∫ ((4x^2 −4x +1 +x^2 +2x +1)/(x^2 +2x +1))dx =∫ ((5x^2 −2x +2)/(x^2 +2x+1))dx =∫ ((5(x^2 +2x+1)−12x−3)/(x^2 +2x +1))dx =5x − ∫ ((12x−3)/(x^2 +2x+1)) dx but ∫ ((12x−3)/(x^2 +2x+1))dx=∫ ((12x−3)/((x+1)^2 ))dx =∫ ((12(x+1)−15)/((x+1)^2 ))dx =12 ∫ (dx/(x+1)) −15 ∫ (dx/((x+1)^2 )) =12ln∣x+1∣+((15)/(x+1)) +c ⇒ ∫ f^(−1) (x)dx =5x −12ln∣x+1∣ −((15)/(x+1)) +c .
$$\left.\mathrm{3}\right)\:\int\:{f}^{−\mathrm{1}} \left({x}\right){dx}\:=\:\int\:\:\left(\:\frac{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}\:+\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{1}}\:+\mathrm{1}\right){dx}=\:\int\:\:\frac{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}\:+\mathrm{1}\:+{x}^{\mathrm{2}} \:+\mathrm{2}{x}\:+\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}\:+\mathrm{1}}{dx} \\ $$$$=\int\:\:\frac{\mathrm{5}{x}^{\mathrm{2}} \:−\mathrm{2}{x}\:+\mathrm{2}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{1}}{dx}\:=\int\:\:\frac{\mathrm{5}\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{1}\right)−\mathrm{12}{x}−\mathrm{3}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}\:+\mathrm{1}}{dx} \\ $$$$=\mathrm{5}{x}\:−\:\int\:\:\frac{\mathrm{12}{x}−\mathrm{3}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{1}}\:{dx}\:{but}\:\:\int\:\:\frac{\mathrm{12}{x}−\mathrm{3}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{1}}{dx}=\int\:\frac{\mathrm{12}{x}−\mathrm{3}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$=\int\:\:\frac{\mathrm{12}\left({x}+\mathrm{1}\right)−\mathrm{15}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:=\mathrm{12}\:\int\:\frac{{dx}}{{x}+\mathrm{1}}\:−\mathrm{15}\:\int\:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:=\mathrm{12}{ln}\mid{x}+\mathrm{1}\mid+\frac{\mathrm{15}}{{x}+\mathrm{1}}\:\:+{c}\:\Rightarrow \\ $$$$\int\:{f}^{−\mathrm{1}} \left({x}\right){dx}\:=\mathrm{5}{x}\:−\mathrm{12}{ln}\mid{x}+\mathrm{1}\mid\:−\frac{\mathrm{15}}{{x}+\mathrm{1}}\:+{c}\:. \\ $$
Commented by maxmathsup by imad last updated on 17/Sep/18
4) ∫ (dx/(f^(−1) (x))) = ∫ ((x^2 +2x+1)/(5x^2 −2x +2)) dx =(1/5) ∫ ((5x^2 +10x +5)/(5x^2 −2x +2))dx =(1/5) ∫ ((5x^2 −2x +2 +12x +3)/(5x^2 −2x +2))dx =(1/5)x +(1/5) ∫ ((12x+3)/(5x^2 −2x +2))dx but ∫ ((12x+3)/(5x^2 −2x +2)) dx = ∫ ((10x−2 +2x +5)/(5x^2 −2x +2))dx =ln∣5x^2 −2x +2∣ +∫ ((2x+5)/(5x^2 −2x +2)) =ln∣5x^2 −2x +2∣ +(1/5) ∫ ((10x −2+27)/(5x^2 −2x +2))dx =(6/5)ln∣5x^2 −2x +2∣ +((27)/(25)) ∫ (dx/(x^2 −(2/5)x +(2/5))) but ∫ (dx/(x^2 −(2/5)x+(2/5))) =∫ (dx/(x^2 −(2/5)x +(1/(25)) +((10)/(25))−(1/(25)))) =∫ (dx/((x−(1/5))^2 +(9/(25)))) =_(x−(1/5)=(3/5)u) ((25)/9)∫ (1/(1+u^2 )) (3/5)du =(5/3) arctan(((5x−1)/3)) +c ⇒ ∫ (dx/(f^(−1) (x))) = (x/5) +(6/(25))ln∣5x^2 −2x+2∣ +((27)/(125)) (5/3) arctan(((5x−1)/3)) +c =(x/5) +(6/(25))ln∣5x^2 −2x +2∣ +(9/(25)) arctan(((5x−1)/3))+c .
$$\left.\mathrm{4}\right)\:\int\:\:\frac{{dx}}{{f}^{−\mathrm{1}} \left({x}\right)}\:=\:\int\:\:\:\frac{{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{1}}{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}\:+\mathrm{2}}\:{dx}\:=\frac{\mathrm{1}}{\mathrm{5}}\:\int\:\:\frac{\mathrm{5}{x}^{\mathrm{2}} \:+\mathrm{10}{x}\:+\mathrm{5}}{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}\:+\mathrm{2}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\:\int\:\:\:\frac{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}\:+\mathrm{2}\:\:+\mathrm{12}{x}\:+\mathrm{3}}{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}\:+\mathrm{2}}{dx}\:=\frac{\mathrm{1}}{\mathrm{5}}{x}\:+\frac{\mathrm{1}}{\mathrm{5}}\:\int\:\:\frac{\mathrm{12}{x}+\mathrm{3}}{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}\:+\mathrm{2}}{dx}\:{but} \\ $$$$\int\:\:\frac{\mathrm{12}{x}+\mathrm{3}}{\mathrm{5}{x}^{\mathrm{2}} \:−\mathrm{2}{x}\:+\mathrm{2}}\:{dx}\:=\:\int\:\:\:\frac{\mathrm{10}{x}−\mathrm{2}\:+\mathrm{2}{x}\:+\mathrm{5}}{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}\:+\mathrm{2}}{dx} \\ $$$$={ln}\mid\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}\:+\mathrm{2}\mid\:+\int\:\:\:\frac{\mathrm{2}{x}+\mathrm{5}}{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}\:+\mathrm{2}} \\ $$$$={ln}\mid\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}\:+\mathrm{2}\mid\:\:+\frac{\mathrm{1}}{\mathrm{5}}\:\int\:\:\:\frac{\mathrm{10}{x}\:−\mathrm{2}+\mathrm{27}}{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}\:+\mathrm{2}}{dx} \\ $$$$=\frac{\mathrm{6}}{\mathrm{5}}{ln}\mid\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}\:+\mathrm{2}\mid\:+\frac{\mathrm{27}}{\mathrm{25}}\:\int\:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:−\frac{\mathrm{2}}{\mathrm{5}}{x}\:\:+\frac{\mathrm{2}}{\mathrm{5}}}\:{but} \\ $$$$\int\:\:\frac{{dx}}{{x}^{\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{5}}{x}+\frac{\mathrm{2}}{\mathrm{5}}}\:=\int\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:−\frac{\mathrm{2}}{\mathrm{5}}{x}\:+\frac{\mathrm{1}}{\mathrm{25}}\:+\frac{\mathrm{10}}{\mathrm{25}}−\frac{\mathrm{1}}{\mathrm{25}}}\:=\int\:\:\:\:\:\frac{{dx}}{\left({x}−\frac{\mathrm{1}}{\mathrm{5}}\right)^{\mathrm{2}} \:+\frac{\mathrm{9}}{\mathrm{25}}} \\ $$$$=_{{x}−\frac{\mathrm{1}}{\mathrm{5}}=\frac{\mathrm{3}}{\mathrm{5}}{u}} \:\:\:\:\frac{\mathrm{25}}{\mathrm{9}}\int\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\frac{\mathrm{3}}{\mathrm{5}}{du}\:=\frac{\mathrm{5}}{\mathrm{3}}\:{arctan}\left(\frac{\mathrm{5}{x}−\mathrm{1}}{\mathrm{3}}\right)\:+{c}\:\Rightarrow \\ $$$$\int\:\:\frac{{dx}}{{f}^{−\mathrm{1}} \left({x}\right)}\:\:\:=\:\:\frac{{x}}{\mathrm{5}}\:+\frac{\mathrm{6}}{\mathrm{25}}{ln}\mid\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\mid\:+\frac{\mathrm{27}}{\mathrm{125}}\:\frac{\mathrm{5}}{\mathrm{3}}\:{arctan}\left(\frac{\mathrm{5}{x}−\mathrm{1}}{\mathrm{3}}\right)\:+{c} \\ $$$$=\frac{{x}}{\mathrm{5}}\:+\frac{\mathrm{6}}{\mathrm{25}}{ln}\mid\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}\:+\mathrm{2}\mid\:+\frac{\mathrm{9}}{\mathrm{25}}\:{arctan}\left(\frac{\mathrm{5}{x}−\mathrm{1}}{\mathrm{3}}\right)+{c}\:. \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *