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$$\int_{a}^{b}{\left(\frac{1}{x^2} +1\right)^{\frac{1}{2}}dx}$$ for $0<a<b$.

I don't know how to compute this at all.

Can you give me a hint please?

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Hint: let x = tan(u), and 1 + (tan(u))^2 = (sec(u))^2, and dx = (sec(u))^2*du, and you can take it from there.

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  • $\begingroup$ I did like that. but I can't find the answer... $\endgroup$ – user114952 Jan 6 '14 at 8:01
  • $\begingroup$ I did it thx a lot $\endgroup$ – user114952 Jan 6 '14 at 8:03
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The integral $$\int_{a}^{b}{\left(\frac{1}{x^2} +1\right)^{\frac{1}{2}}dx}$$ when $0<a<b$ can be written as $$\int_{a}^{b}{\frac{\sqrt{1+x^2}}{x}dx}=\int_a^bx^{-1}\sqrt{1+x^2}dx$$ Now set $1+x^2=t^2$.

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