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Question Number 12377 by tawa last updated on 20/Apr/17
∫ ((√(1 + (√x)))/x) dx
$$\int\:\:\frac{\sqrt{\mathrm{1}\:+\:\sqrt{\mathrm{x}}}}{\mathrm{x}}\:\:\mathrm{dx} \\ $$
Answered by mrW1 last updated on 21/Apr/17
t^2 =(√x) t^4 =x 4t^3 dt=dx I=∫((√(1+(√x)))/x)dx=∫((√(1+t^2 ))/t^4 )×4t^3 dt =4∫((√(1+t^2 ))/t)dt=2∫((√(1+t^2 ))/t^2 )d(t^2 ) =2∫((√(1+s))/s)ds (s=t^2 =(√x)) ∫((√(1+s))/s)ds=2(√(1+s))+ln (((√(1+s))−1)/( (√(1+s))+1))+C I=4(√(1+s))+2ln (((√(1+s))−1)/( (√(1+s))+1))+C =4(√(1+(√x)))+2ln (((√(1+(√x)))−1)/( (√(1+(√x)))+1))+C
$${t}^{\mathrm{2}} =\sqrt{{x}} \\ $$$${t}^{\mathrm{4}} ={x} \\ $$$$\mathrm{4}{t}^{\mathrm{3}} {dt}={dx} \\ $$$${I}=\int\frac{\sqrt{\mathrm{1}+\sqrt{{x}}}}{{x}}{dx}=\int\frac{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{{t}^{\mathrm{4}} }×\mathrm{4}{t}^{\mathrm{3}} {dt} \\ $$$$=\mathrm{4}\int\frac{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{{t}}{dt}=\mathrm{2}\int\frac{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{{t}^{\mathrm{2}} }{d}\left({t}^{\mathrm{2}} \right) \\ $$$$=\mathrm{2}\int\frac{\sqrt{\mathrm{1}+{s}}}{{s}}{ds}\:\:\:\:\:\:\:\left({s}={t}^{\mathrm{2}} =\sqrt{{x}}\right) \\ $$$$ \\ $$$$\int\frac{\sqrt{\mathrm{1}+{s}}}{{s}}{ds}=\mathrm{2}\sqrt{\mathrm{1}+{s}}+\mathrm{ln}\:\frac{\sqrt{\mathrm{1}+{s}}−\mathrm{1}}{\:\sqrt{\mathrm{1}+{s}}+\mathrm{1}}+{C} \\ $$$$ \\ $$$${I}=\mathrm{4}\sqrt{\mathrm{1}+{s}}+\mathrm{2ln}\:\frac{\sqrt{\mathrm{1}+{s}}−\mathrm{1}}{\:\sqrt{\mathrm{1}+{s}}+\mathrm{1}}+{C} \\ $$$$=\mathrm{4}\sqrt{\mathrm{1}+\sqrt{{x}}}+\mathrm{2ln}\:\frac{\sqrt{\mathrm{1}+\sqrt{{x}}}−\mathrm{1}}{\:\sqrt{\mathrm{1}+\sqrt{{x}}}+\mathrm{1}}+{C} \\ $$
Commented by tawa last updated on 21/Apr/17
wow. God bless you sir.
$$\mathrm{wow}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by ajfour last updated on 21/Apr/17
let x=tan^4 t⇒ dx=4tan^3 tsec^2 t ∫((√(1+(√x)))/x)dx =∫(((sec t)(4tan^3 tsec^2 t))/(tan^4 t))dt =4∫cosec tsec^2 tdt =4[cosec ttan t−∫tan t(−cosec tcot t)dt] =4[(1/(cos t)) + ∫cosec tdt]+c =4(√(1+(√x))) +4ln ∣(√(1+(1/( (√x))))) −(√(1/( (√x)))) ∣+c′ = 4(√(1+(√x))) +4ln ∣(√(1+(√x))) −1∣−ln ∣x∣+c′ .
$${let}\:\:{x}=\mathrm{tan}\:^{\mathrm{4}} {t}\Rightarrow\:{dx}=\mathrm{4tan}\:^{\mathrm{3}} {t}\mathrm{sec}\:^{\mathrm{2}} {t} \\ $$$$\int\frac{\sqrt{\mathrm{1}+\sqrt{{x}}}}{{x}}{dx}\:=\int\frac{\left(\mathrm{sec}\:{t}\right)\left(\mathrm{4tan}\:^{\mathrm{3}} {t}\mathrm{sec}\:^{\mathrm{2}} {t}\right)}{\mathrm{tan}\:^{\mathrm{4}} {t}}{dt} \\ $$$$=\mathrm{4}\int\mathrm{cosec}\:{t}\mathrm{sec}\:^{\mathrm{2}} {tdt} \\ $$$$=\mathrm{4}\left[\mathrm{cosec}\:{t}\mathrm{tan}\:{t}−\int\mathrm{tan}\:{t}\left(−\mathrm{cosec}\:{t}\mathrm{cot}\:{t}\right){dt}\right] \\ $$$$=\mathrm{4}\left[\frac{\mathrm{1}}{\mathrm{cos}\:{t}}\:+\:\int\mathrm{cosec}\:{tdt}\right]+{c} \\ $$$$=\mathrm{4}\sqrt{\mathrm{1}+\sqrt{{x}}}\:+\mathrm{4ln}\:\mid\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{x}}}}\:−\sqrt{\frac{\mathrm{1}}{\:\sqrt{{x}}}}\:\mid+{c}' \\ $$$$=\:\mathrm{4}\sqrt{\mathrm{1}+\sqrt{{x}}}\:+\mathrm{4ln}\:\mid\sqrt{\mathrm{1}+\sqrt{{x}}}\:−\mathrm{1}\mid−\mathrm{ln}\:\mid{x}\mid+{c}'\:. \\ $$
Commented by tawa last updated on 21/Apr/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 21/Apr/17
x=cos^2 θ⇒dx=−2sinθcosθdθ I=∫((√(1+cosθ))/(cos^2 θ))dθ=(√2)∫((cos(θ/2))/(cos^2 θ))(−2sinθcosθ)dθ =−4(√2)∫((cos^2 (θ/2)sin(θ/2))/(2cos^2 (θ/2)−1))dθ=8(√2)∫((ϕ^2 dϕ)/(2ϕ^2 −1))= 4(√2)∫(((2ϕ^2 −1)+1)/(2ϕ^2 −1))dϕ=4(√2)∫(1−(1/(2ϕ^2 −1)))dϕ= =4(√2)[ϕ−(1/2)∫(((((√2)ϕ−1)−((√2)ϕ+1)))/(((√2)ϕ−1)((√2)ϕ+1)))dϕ]= =4(√2)[ϕ−(1/2)∫((1/( (√2)ϕ+1))−(1/( (√2)ϕ−1)))dϕ]= =4(√2)[ϕ−(1/(2(√2)))ln(((√2)ϕ+1)/( (√2)ϕ−1))]+C= =4(√2)cos(θ/2)−2ln(((√2)cos(θ/2)+1)/( (√2)cos(θ/2)−1))+C cos^2 θ=x⇒cosθ=(√x)⇒2cos^2 (θ/2)=1+(√x) cos(θ/2)=((√(1+(√x)))/( (√2))) I=4(√2)((√(1+(√x)))/( (√2)))−2ln(((√2)((√(1+(√x)))/( (√2)))+1)/( (√2)((√(1+(√x)))/( (√2)))−1))+C= I=4(√(1+(√x)))−2ln((((√(1+(√x)))+1)/( (√(1+(√x)))−1)))+C .■
$${x}={cos}^{\mathrm{2}} \theta\Rightarrow{dx}=−\mathrm{2}{sin}\theta{cos}\theta{d}\theta \\ $$$${I}=\int\frac{\sqrt{\mathrm{1}+{cos}\theta}}{{cos}^{\mathrm{2}} \theta}{d}\theta=\sqrt{\mathrm{2}}\int\frac{{cos}\frac{\theta}{\mathrm{2}}}{{cos}^{\mathrm{2}} \theta}\left(−\mathrm{2}{sin}\theta{cos}\theta\right){d}\theta \\ $$$$=−\mathrm{4}\sqrt{\mathrm{2}}\int\frac{{cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}{sin}\frac{\theta}{\mathrm{2}}}{\mathrm{2}{cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}−\mathrm{1}}{d}\theta=\mathrm{8}\sqrt{\mathrm{2}}\int\frac{\varphi^{\mathrm{2}} {d}\varphi}{\mathrm{2}\varphi^{\mathrm{2}} −\mathrm{1}}= \\ $$$$\mathrm{4}\sqrt{\mathrm{2}}\int\frac{\left(\mathrm{2}\varphi^{\mathrm{2}} −\mathrm{1}\right)+\mathrm{1}}{\mathrm{2}\varphi^{\mathrm{2}} −\mathrm{1}}{d}\varphi=\mathrm{4}\sqrt{\mathrm{2}}\int\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}\varphi^{\mathrm{2}} −\mathrm{1}}\right){d}\varphi= \\ $$$$=\mathrm{4}\sqrt{\mathrm{2}}\left[\varphi−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\left(\left(\sqrt{\mathrm{2}}\varphi−\mathrm{1}\right)−\left(\sqrt{\mathrm{2}}\varphi+\mathrm{1}\right)\right)}{\left(\sqrt{\mathrm{2}}\varphi−\mathrm{1}\right)\left(\sqrt{\mathrm{2}}\varphi+\mathrm{1}\right)}{d}\varphi\right]= \\ $$$$=\mathrm{4}\sqrt{\mathrm{2}}\left[\varphi−\frac{\mathrm{1}}{\mathrm{2}}\int\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\varphi+\mathrm{1}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\varphi−\mathrm{1}}\right){d}\varphi\right]= \\ $$$$=\mathrm{4}\sqrt{\mathrm{2}}\left[\varphi−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\frac{\sqrt{\mathrm{2}}\varphi+\mathrm{1}}{\:\sqrt{\mathrm{2}}\varphi−\mathrm{1}}\right]+{C}= \\ $$$$=\mathrm{4}\sqrt{\mathrm{2}}{cos}\frac{\theta}{\mathrm{2}}−\mathrm{2}{ln}\frac{\sqrt{\mathrm{2}}{cos}\frac{\theta}{\mathrm{2}}+\mathrm{1}}{\:\sqrt{\mathrm{2}}{cos}\frac{\theta}{\mathrm{2}}−\mathrm{1}}+{C} \\ $$$${cos}^{\mathrm{2}} \theta={x}\Rightarrow{cos}\theta=\sqrt{{x}}\Rightarrow\mathrm{2}{cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}=\mathrm{1}+\sqrt{{x}} \\ $$$${cos}\frac{\theta}{\mathrm{2}}=\frac{\sqrt{\mathrm{1}+\sqrt{\boldsymbol{{x}}}}}{\:\sqrt{\mathrm{2}}} \\ $$$$\boldsymbol{{I}}=\mathrm{4}\sqrt{\mathrm{2}}\frac{\sqrt{\mathrm{1}+\sqrt{\boldsymbol{{x}}}}}{\:\sqrt{\mathrm{2}}}−\mathrm{2}\boldsymbol{{ln}}\frac{\sqrt{\mathrm{2}}\frac{\sqrt{\mathrm{1}+\sqrt{\boldsymbol{{x}}}}}{\:\sqrt{\mathrm{2}}}+\mathrm{1}}{\:\sqrt{\mathrm{2}}\frac{\sqrt{\mathrm{1}+\sqrt{\boldsymbol{{x}}}}}{\:\sqrt{\mathrm{2}}}−\mathrm{1}}+\boldsymbol{{C}}= \\ $$$${I}=\mathrm{4}\sqrt{\mathrm{1}+\sqrt{\boldsymbol{{x}}}}−\mathrm{2}\boldsymbol{{ln}}\left(\frac{\sqrt{\mathrm{1}+\sqrt{\boldsymbol{{x}}}}+\mathrm{1}}{\:\sqrt{\mathrm{1}+\sqrt{\boldsymbol{{x}}}}−\mathrm{1}}\right)+\boldsymbol{{C}}\:.\blacksquare \\ $$
Commented by tawa last updated on 21/Apr/17
i really appreciate sir.
$$\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$
Commented by tawa last updated on 21/Apr/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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