Question Number 116813 by bemath last updated on 07/Oct/20
$$\:\:\:\:\:\:\underset{\mathrm{1}} {\overset{\sqrt{\mathrm{3}}} {\int}}\:\frac{\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}{\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}\:? \\ $$
Answered by bobhans last updated on 07/Oct/20
![letting x = tan θ → { ((θ=(π/3))),((θ=(π/4))) :} I=∫_(π/4) ^(π/3) (((√(1+tan^2 θ)) sec^2 θ)/(tan^2 θ)) dθ I=∫_(π/4) ^(π/3) ((sec θ (tan^2 θ+1) )/(tan^2 θ))dθ I=∫_(π/4) ^(π/3) (sec θ + ((sec θ)/(tan ^2 θ)))dθ I= ln ∣sec θ+tan θ∣_(π/4) ^(π/3) + I_2 I_2 = ∫((sec θ)/(tan ^2 θ)) dθ = ∫ ((cos θ)/(sin^2 θ)) dθ = ∫ ((d(sin θ))/(sin^2 θ)) I_2 =[(1/(sin θ)) ]_(π/4) ^(π/3) = [(2/( (√3))) −(2/( (√2))) ]=(2/( (√3)))−(2/( (√2))) therefore I=ln ∣((2+(√3))/( (√2)+1)) ∣+((2((√2)−(√3)))/( (√6)))](https://www.tinkutara.com/question/Q116814.png)
$$\:\mathrm{letting}\:\mathrm{x}\:=\:\mathrm{tan}\:\theta\:\rightarrow\begin{cases}{\theta=\frac{\pi}{\mathrm{3}}}\\{\theta=\frac{\pi}{\mathrm{4}}}\end{cases} \\ $$$$\mathrm{I}=\int_{\pi/\mathrm{4}} ^{\pi/\mathrm{3}} \:\frac{\sqrt{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \theta}\:\mathrm{sec}\:^{\mathrm{2}} \theta}{\mathrm{tan}\:^{\mathrm{2}} \:\theta}\:\mathrm{d}\theta \\ $$$$\mathrm{I}=\int_{\pi/\mathrm{4}} ^{\pi/\mathrm{3}} \:\frac{\mathrm{sec}\:\theta\:\left(\mathrm{tan}^{\mathrm{2}} \:\theta+\mathrm{1}\right)\:}{\mathrm{tan}\:^{\mathrm{2}} \theta}\mathrm{d}\theta \\ $$$$\mathrm{I}=\int_{\pi/\mathrm{4}} ^{\pi/\mathrm{3}} \left(\mathrm{sec}\:\theta\:+\:\frac{\mathrm{sec}\:\theta}{\mathrm{tan}\:\:^{\mathrm{2}} \theta}\right)\mathrm{d}\theta \\ $$$$\mathrm{I}=\:\mathrm{ln}\:\mid\mathrm{sec}\:\theta+\mathrm{tan}\:\theta\mid_{\pi/\mathrm{4}} ^{\pi/\mathrm{3}} \:+\:\mathrm{I}_{\mathrm{2}} \\ $$$$\mathrm{I}_{\mathrm{2}} \:=\:\int\frac{\mathrm{sec}\:\theta}{\mathrm{tan}\:\:^{\mathrm{2}} \theta}\:\mathrm{d}\theta\:=\:\int\:\frac{\mathrm{cos}\:\theta}{\mathrm{sin}\:^{\mathrm{2}} \theta}\:\mathrm{d}\theta\:=\:\int\:\frac{\mathrm{d}\left(\mathrm{sin}\:\theta\right)}{\mathrm{sin}\:^{\mathrm{2}} \theta} \\ $$$$\mathrm{I}_{\mathrm{2}} =\left[\frac{\mathrm{1}}{\mathrm{sin}\:\theta}\:\right]_{\pi/\mathrm{4}} ^{\pi/\mathrm{3}} \:=\:\left[\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:−\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}}\:\right]=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}} \\ $$$$\mathrm{therefore}\:\mathrm{I}=\mathrm{ln}\:\mid\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{2}}+\mathrm{1}}\:\mid+\frac{\mathrm{2}\left(\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}\right)}{\:\sqrt{\mathrm{6}}} \\ $$
Answered by Bird last updated on 07/Oct/20
![I =∫_1 ^(√3) ((√(1+x^2 ))/x^2 )dx by parts I =[−(1/x)(√(1+x^2 ))]_1 ^(√3) +∫_1 ^(√3) (1/x)×(x/( (√(1+x^2 ))))dx =(√2)−(1/( (√3)))(2)+∫_1 ^(√3) (dx/( (√(1+x^2 )))) =(√2)−(2/( (√3))) +[ln(x+(√(1+x^2 )))]_1 ^(√3) =(√2)−(2/( (√3))) +{ln((√3)+2)−ln(1+(√2))} =(√2)−((2(√3))/3) +ln(((2+(√3))/(1+(√2))))](https://www.tinkutara.com/question/Q116867.png)
$${I}\:=\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \:\:\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} }{dx}\:\:{by}\:{parts} \\ $$$${I}\:=\left[−\frac{\mathrm{1}}{{x}}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right]_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} +\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \frac{\mathrm{1}}{{x}}×\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx} \\ $$$$=\sqrt{\mathrm{2}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left(\mathrm{2}\right)+\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \frac{{dx}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$$=\sqrt{\mathrm{2}}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:+\left[{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\right]_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \\ $$$$=\sqrt{\mathrm{2}}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:+\left\{{ln}\left(\sqrt{\mathrm{3}}+\mathrm{2}\right)−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\right\} \\ $$$$=\sqrt{\mathrm{2}}−\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\:+{ln}\left(\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{1}+\sqrt{\mathrm{2}}}\right) \\ $$