Question Number 37004 by behi83417@gmail.com last updated on 07/Jun/18
Answered by tanmay.chaudhury50@gmail.com last updated on 07/Jun/18
$$\int\frac{\sqrt{\mathrm{1}+{x}^{{p}} \:\:}}{{x}}{dx} \\ $$$$\int\frac{\mathrm{1}+{x}^{{p}} }{{x}\sqrt{\mathrm{1}+{x}^{{p}} \:}}{dx} \\ $$$$\int\frac{{x}^{{p}−\mathrm{1}} \left(\mathrm{1}+{x}^{{p}} \right)}{{x}^{{p}} \sqrt{\mathrm{1}+{x}^{{p}} \:}}{dx} \\ $$$${t}={x}^{{p}} \\ $$$${dt}={px}^{{p}−\mathrm{1}} {dx} \\ $$$$\int\frac{{dt}}{{p}}×\frac{\left(\mathrm{1}+{t}\right)}{{t}\sqrt{\mathrm{1}+{t}}} \\ $$$$\frac{\mathrm{1}}{{p}}\int\frac{{dt}}{{t}\sqrt{\mathrm{1}+{t}}}\:+\frac{\mathrm{1}}{{p}}\int\frac{{dt}}{\:\sqrt{\mathrm{1}+{t}}} \\ $$$${k}^{\mathrm{2}} =\mathrm{1}+{t}\:\:\:\mathrm{2}{kdk}={dt} \\ $$$$\frac{\mathrm{1}}{{p}}\int\frac{\mathrm{2}{kdk}}{\left({k}^{\mathrm{2}} −\mathrm{1}\right){k}}+\frac{\mathrm{1}}{{p}}\int\frac{\mathrm{2}{kdk}}{{k}} \\ $$$$\frac{\mathrm{2}}{{p}}\int\frac{{dk}}{{k}^{\mathrm{2}} −\mathrm{1}}+\frac{\mathrm{2}}{{p}}\int{dk} \\ $$$$\frac{\mathrm{2}}{{p}}×\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\left({k}+\mathrm{1}\right)−\left({k}−\mathrm{1}\right)}{\left({k}+\mathrm{1}\right)\left({k}−\mathrm{1}\right)}{dk}+\frac{\mathrm{2}}{{p}}\int{dk} \\ $$$$\frac{\mathrm{1}}{{p}}{ln}\mid\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}}\mid+\frac{\mathrm{2}}{{p}}{k}\:\:+{c} \\ $$$$\frac{\mathrm{1}}{{p}}{ln}\mid\frac{\sqrt{\mathrm{1}+{x}^{{p}} \:\:}\:\:−\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{{p}} \:}\:\:+\mathrm{1}}\mid\:\:+\frac{\mathrm{2}}{{p}}{k} \\ $$$$\frac{\mathrm{1}}{{p}}{ln}\mid\frac{\sqrt{\mathrm{1}+{x}^{{p}} }\:\:−\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}_{} ^{{p}} \:}\:\:+\mathrm{1}}\mid+\frac{\mathrm{2}}{{p}}\sqrt{\mathrm{1}+{x}^{{p}} } \\ $$
Commented by behi83417@gmail.com last updated on 07/Jun/18
$${mr}\:{tanmay}!\:{nice}\:{work}\:{done}\:{by}\:{you}. \\ $$$${thanks}! \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jun/18
$${its}\:{alright}…{thank}\:{you}\:{sir}\: \\ $$
Answered by ajfour last updated on 07/Jun/18
![let x^( p) =tan^2 θ ⇒ px^( p−1) dx=2tan θsec^2 θdθ ⇒ px^p ( (dx/x))=2tan θsec^2 θdθ or ptan^2 θ( (dx/x))=2tan θsec^2 θdθ I=∫((sec θ(2tan θsec^2 θ)dθ)/(ptan^2 θ)) =(2/p)∫(dθ/(sin θcos^2 θ)) =(2/p)∫((sin θdθ)/(cos^2 θ(1−cos^2 θ))) =(2/p)∫(dt/(t^2 (t^2 −1))) (t=cos θ) (1/(t^2 (t−1)(t+1)))=(A/t)+(B/t^2 )+(C/(t−1))+(D/(t+1)) C=(1/2), D=−(1/2) , B=−1 coeff. of t^3 is A+C+D=0 ⇒ A=0 ∴ I= (2/p)[(1/t)+(1/2)ln ∣((t−1)/(t+1))∣]+c As x^( p) = tan^2 θ= sec^2 θ−1=(1/t^2 )−1 ⇒ t=(1/( (√(1+x^p )))) I=(2/p)[(√(1+x^( p) ))+(1/2)ln ∣((1−(√(1+x^p )))/(1+(√(1+x^p ))))∣]+c . for p=1 I_1 = 2(√(1+x^( p) ))+ln ∣((1−(√(1+x)))/(1+(√(1+x))))∣+c .](https://www.tinkutara.com/question/Q37011.png)
$${let}\:\:{x}^{\:{p}} =\mathrm{tan}\:^{\mathrm{2}} \theta \\ $$$$\Rightarrow\:\:\:\:{px}^{\:{p}−\mathrm{1}} {dx}=\mathrm{2tan}\:\theta\mathrm{sec}\:^{\mathrm{2}} \theta{d}\theta \\ $$$$\Rightarrow\:\:\:\:\:\:\:{px}^{{p}} \left(\:\frac{{dx}}{{x}}\right)=\mathrm{2tan}\:\theta\mathrm{sec}\:^{\mathrm{2}} \theta{d}\theta \\ $$$${or}\:\:\:\:\:\:\:{p}\mathrm{tan}\:^{\mathrm{2}} \theta\left(\:\frac{{dx}}{{x}}\right)=\mathrm{2tan}\:\theta\mathrm{sec}\:^{\mathrm{2}} \theta{d}\theta \\ $$$${I}=\int\frac{\mathrm{sec}\:\theta\left(\mathrm{2tan}\:\theta\mathrm{sec}\:^{\mathrm{2}} \theta\right){d}\theta}{{p}\mathrm{tan}\:^{\mathrm{2}} \theta} \\ $$$$\:\:\:\:=\frac{\mathrm{2}}{{p}}\int\frac{{d}\theta}{\mathrm{sin}\:\theta\mathrm{cos}\:^{\mathrm{2}} \theta} \\ $$$$\:\:\:=\frac{\mathrm{2}}{{p}}\int\frac{\mathrm{sin}\:\theta{d}\theta}{\mathrm{cos}\:^{\mathrm{2}} \theta\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \theta\right)} \\ $$$$\:\:\:=\frac{\mathrm{2}}{{p}}\int\frac{{dt}}{{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{1}\right)}\:\:\:\:\:\:\:\:\left({t}=\mathrm{cos}\:\theta\right) \\ $$$$\:\:\frac{\mathrm{1}}{{t}^{\mathrm{2}} \left({t}−\mathrm{1}\right)\left({t}+\mathrm{1}\right)}=\frac{{A}}{{t}}+\frac{{B}}{{t}^{\mathrm{2}} }+\frac{{C}}{{t}−\mathrm{1}}+\frac{{D}}{{t}+\mathrm{1}} \\ $$$$\:\:{C}=\frac{\mathrm{1}}{\mathrm{2}},\:\:{D}=−\frac{\mathrm{1}}{\mathrm{2}}\:,\:\:{B}=−\mathrm{1} \\ $$$${coeff}.\:{of}\:{t}^{\mathrm{3}} \:{is}\:\:{A}+{C}+{D}=\mathrm{0}\: \\ $$$$\Rightarrow\:\:\:{A}=\mathrm{0} \\ $$$$\therefore\:{I}=\:\frac{\mathrm{2}}{{p}}\left[\frac{\mathrm{1}}{{t}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}\mid\right]+{c}\: \\ $$$${As}\:\:{x}^{\:{p}} \:=\:\mathrm{tan}\:^{\mathrm{2}} \theta=\:\mathrm{sec}\:^{\mathrm{2}} \theta−\mathrm{1}=\frac{\mathrm{1}}{{t}^{\mathrm{2}} }−\mathrm{1} \\ $$$$\Rightarrow\:{t}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{{p}} }} \\ $$$${I}=\frac{\mathrm{2}}{{p}}\left[\sqrt{\mathrm{1}+{x}^{\:{p}} }+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid\frac{\mathrm{1}−\sqrt{\mathrm{1}+{x}^{{p}} }}{\mathrm{1}+\sqrt{\mathrm{1}+{x}^{{p}} }}\mid\right]+{c}\:. \\ $$$${for}\:\:{p}=\mathrm{1} \\ $$$$\:\:\:\:\:{I}_{\mathrm{1}} =\:\mathrm{2}\sqrt{\mathrm{1}+{x}^{\:{p}} }+\mathrm{ln}\:\mid\frac{\mathrm{1}−\sqrt{\mathrm{1}+{x}}}{\mathrm{1}+\sqrt{\mathrm{1}+{x}}}\mid+{c}\:. \\ $$
Commented by behi83417@gmail.com last updated on 07/Jun/18
$${thanks}\:{in}\:{advance}\:{sir}\:{Ajfour}! \\ $$$${perfect}\:{method}. \\ $$
Answered by MJS last updated on 07/Jun/18
![p=0 (√2)∫(1/x)dx=(√2)ln x +C p≠0 ∫((√(1+x^p ))/x)dx= [t=x^p → dx=(x^(1−p) /p)dt] =(1/p)∫((√(1+t))/t)dt= [u=(√(1+t)) → dt=2(√(1+t))du] =(2/p)∫(u^2 /(u^2 −1))du=(2/p)(∫du+∫(du/(u^2 −1)))= =(2/p)(u+(1/2)(∫(du/(u−1))−∫(du/(u+1))))= =(2/p)(u+(1/2)(ln(u−1)−ln(u+1)))= =(2/p)(u+(1/2)ln ((u−1)/(u+1)))=(2/p)(√(1+t))+(1/p)ln (((√(1+t))−1)/( (√(1+t))+1))= =(2/p)(√(1+x^p ))+(1/p)ln (((√(1+x^p ))−1)/( (√(1+x^p ))+1))+C](https://www.tinkutara.com/question/Q37012.png)
$${p}=\mathrm{0} \\ $$$$\sqrt{\mathrm{2}}\int\frac{\mathrm{1}}{{x}}{dx}=\sqrt{\mathrm{2}}\mathrm{ln}\:{x}\:+{C} \\ $$$${p}\neq\mathrm{0} \\ $$$$\int\frac{\sqrt{\mathrm{1}+{x}^{{p}} }}{{x}}{dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{t}={x}^{{p}} \:\rightarrow\:{dx}=\frac{{x}^{\mathrm{1}−{p}} }{{p}}{dt}\right] \\ $$$$=\frac{\mathrm{1}}{{p}}\int\frac{\sqrt{\mathrm{1}+{t}}}{{t}}{dt}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{u}=\sqrt{\mathrm{1}+{t}}\:\rightarrow\:{dt}=\mathrm{2}\sqrt{\mathrm{1}+{t}}{du}\right] \\ $$$$=\frac{\mathrm{2}}{{p}}\int\frac{{u}^{\mathrm{2}} }{{u}^{\mathrm{2}} −\mathrm{1}}{du}=\frac{\mathrm{2}}{{p}}\left(\int{du}+\int\frac{{du}}{{u}^{\mathrm{2}} −\mathrm{1}}\right)= \\ $$$$=\frac{\mathrm{2}}{{p}}\left({u}+\frac{\mathrm{1}}{\mathrm{2}}\left(\int\frac{{du}}{{u}−\mathrm{1}}−\int\frac{{du}}{{u}+\mathrm{1}}\right)\right)= \\ $$$$=\frac{\mathrm{2}}{{p}}\left({u}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{ln}\left({u}−\mathrm{1}\right)−\mathrm{ln}\left({u}+\mathrm{1}\right)\right)\right)= \\ $$$$=\frac{\mathrm{2}}{{p}}\left({u}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\frac{{u}−\mathrm{1}}{{u}+\mathrm{1}}\right)=\frac{\mathrm{2}}{{p}}\sqrt{\mathrm{1}+{t}}+\frac{\mathrm{1}}{{p}}\mathrm{ln}\:\frac{\sqrt{\mathrm{1}+{t}}−\mathrm{1}}{\:\sqrt{\mathrm{1}+{t}}+\mathrm{1}}= \\ $$$$=\frac{\mathrm{2}}{{p}}\sqrt{\mathrm{1}+{x}^{{p}} }+\frac{\mathrm{1}}{{p}}\mathrm{ln}\:\frac{\sqrt{\mathrm{1}+{x}^{{p}} }−\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{{p}} }+\mathrm{1}}+{C} \\ $$
Commented by behi83417@gmail.com last updated on 07/Jun/18
$${thank}\:{you}\:{very}\:{much}\:{sir}! \\ $$$${simple}\:{and}\:{smart}. \\ $$