Question Number 45043 by maxmathsup by imad last updated on 07/Oct/18
$${let}\:{f}\left({x}\right)=\frac{\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{{x}} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\:\int_{\mathrm{1}} ^{\mathrm{3}} {f}\left({x}\right){dx} \\ $$$$\left.\mathrm{2}\right)\:{determine}\:{f}^{−\mathrm{1}} \left({x}\right) \\ $$$$\left.\mathrm{3}\right)\:{find}\:\int\:{f}^{−\mathrm{1}} \left({x}\right){dx}\:\:. \\ $$
Answered by MJS last updated on 08/Oct/18
![∫((1+(√(x^2 +1)))/x)dx= [t=(√(x^2 +1)) → dx=((√(x^2 +1))/x)dt] =∫(t/(t−1))dt= [u=t−1 → dt=du] =∫((u+1)/u)du=u+ln u=t−1+ln(t−1)= =(√(x^2 +1))−1+ln∣(√(x^2 +1))−1∣= =ln∣1−(√(x^2 +1))∣+(√(x^2 +1))+C (1) ∫_1 ^3 f(x)dx=ln((√(10))+2(√5)−(√2)−1) +((√5)−1)(√2)≈3.40060](https://www.tinkutara.com/question/Q45098.png)
$$\int\frac{\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:\rightarrow\:{dx}=\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{x}}{dt}\right] \\ $$$$=\int\frac{{t}}{{t}−\mathrm{1}}{dt}= \\ $$$$\:\:\:\:\:\left[{u}={t}−\mathrm{1}\:\rightarrow\:{dt}={du}\right] \\ $$$$=\int\frac{{u}+\mathrm{1}}{{u}}{du}={u}+\mathrm{ln}\:{u}={t}−\mathrm{1}+\mathrm{ln}\left({t}−\mathrm{1}\right)= \\ $$$$=\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−\mathrm{1}+\mathrm{ln}\mid\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−\mathrm{1}\mid= \\ $$$$=\mathrm{ln}\mid\mathrm{1}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\mid+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}+{C} \\ $$$$\left(\mathrm{1}\right)\:\underset{\mathrm{1}} {\overset{\mathrm{3}} {\int}}{f}\left({x}\right){dx}=\mathrm{ln}\left(\sqrt{\mathrm{10}}+\mathrm{2}\sqrt{\mathrm{5}}−\sqrt{\mathrm{2}}−\mathrm{1}\right)\:+\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)\sqrt{\mathrm{2}}\approx\mathrm{3}.\mathrm{40060} \\ $$
Commented by maxmathsup by imad last updated on 08/Oct/18
$${thank}\:{you}\:{sir}\:{MJS}. \\ $$
Answered by MJS last updated on 08/Oct/18
![(2) f(x) is defined for x≠0 range is R\[−1; 1] solving x=((1+(√(y^2 +1)))/y) gives y=((2x)/(x^2 −1)) but this is defined for x∈R\{−1; 1} especially for x∈]−1; 1[ ⇒ f^(−1) (x)=((2x)/(x^2 −1)) with x∈R\[−1; 1]](https://www.tinkutara.com/question/Q45101.png)
$$\left(\mathrm{2}\right) \\ $$$${f}\left({x}\right)\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:{x}\neq\mathrm{0} \\ $$$$\mathrm{range}\:\mathrm{is}\:\mathbb{R}\backslash\left[−\mathrm{1};\:\mathrm{1}\right] \\ $$$$\mathrm{solving}\:{x}=\frac{\mathrm{1}+\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}}{{y}}\:\mathrm{gives}\:{y}=\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} −\mathrm{1}}\:\mathrm{but}\:\mathrm{this} \\ $$$$\left.\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{R}\backslash\left\{−\mathrm{1};\:\mathrm{1}\right\}\:\mathrm{especially}\:\mathrm{for}\:{x}\in\right]−\mathrm{1};\:\mathrm{1}\left[\right. \\ $$$$\Rightarrow\:{f}^{−\mathrm{1}} \left({x}\right)=\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} −\mathrm{1}}\:\mathrm{with}\:{x}\in\mathbb{R}\backslash\left[−\mathrm{1};\:\mathrm{1}\right] \\ $$
Answered by MJS last updated on 08/Oct/18
![(3) ∫((2x)/(x^2 −1))dx= [t=x^2 −1 → dx=(dt/(2x))] =∫(dt/t)=ln t =ln ∣x^2 −1∣+C](https://www.tinkutara.com/question/Q45102.png)
$$\left(\mathrm{3}\right) \\ $$$$\int\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} −\mathrm{1}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}={x}^{\mathrm{2}} −\mathrm{1}\:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{2}{x}}\right] \\ $$$$=\int\frac{{dt}}{{t}}=\mathrm{ln}\:{t}\:=\mathrm{ln}\:\mid{x}^{\mathrm{2}} −\mathrm{1}\mid+{C} \\ $$