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$$\int_1^2 \left(1+\frac{1}{2x^2}\right)^{\frac12}dx$$ I have been trying this definite integral,but after many attempts i am stuck if the middle part.I have tried substitution and integration by parts but its not working

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    $\begingroup$ When you get a moment, could you please confirm whether the edit maintained your meaning? That is, should the $x^{2}$ in the radical be in the denominator or not? Both integrals are doable by trig substitution, but the details differ. :) $\endgroup$ – Andrew D. Hwang Aug 3 '17 at 16:06
  • $\begingroup$ The above mentioned ques is right something went wrong while editing $\endgroup$ – Pranab Singh Aug 3 '17 at 16:08
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    $\begingroup$ Is it supposed to be $$\int_1^2\sqrt{1+\dfrac12x^2}\,dx?$$ If so, I would go with the substitution $x=\sqrt2\sinh t$ because $\sqrt{1+\sinh^2t}=\cosh t$. $\endgroup$ – Jyrki Lahtonen Aug 3 '17 at 16:15
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    $\begingroup$ I didn't understand. Are you paying somebody a fixed rate per character for typing here? $\endgroup$ – Jyrki Lahtonen Aug 3 '17 at 16:20
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    $\begingroup$ The standard approach for an integral containing $\sqrt{2x^2+1}$ (as this one does after some easy rewriting of the integrand) is to use the change of variables $x = \frac1{\sqrt2} \tan\theta$. $\endgroup$ – Greg Martin Aug 3 '17 at 17:00
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Substitute $x= \dfrac{\sinh t}{\sqrt{2}}$

Integration extremes become $x=1\to t=\sinh ^{-1}\sqrt{2};\;x=2\to t=\sinh ^{-1}2\sqrt{2}$

$dx=\dfrac{\cosh t}{\sqrt{2}}$

And solve $\dfrac{1}{\sqrt 2}\int \dfrac{\cosh^2 t }{\sinh t} \, dt=\cosh (t)+\log \tanh \dfrac{t}{2}+C$

Can be useful remember

$\sinh t=\dfrac{2 \tanh \frac{t}{2}}{1-\tanh ^2\frac{t}{2}}$

$\cosh (t)=\dfrac{1+\tanh ^2\frac{t}{2}}{1-\tanh ^2\frac{t}{2}}$

And substitute again $\tanh \frac{t}{2}=u \to t= 2 \tanh ^{-1}(u)$ and $dt=\dfrac{2 du}{1-u^2}$

Hope this helps

PS the result is $I=\dfrac{3-\sqrt{3}-\log \left(\sqrt{3}-1\right)}{\sqrt{2}}$

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