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How to evaluate integral $$\int\frac{\sqrt{x^2-1}+x}{\sqrt{x^2-1}+x-1}dx?$$ I tried substitution $u^2=x^2-1$ and $u=\sqrt{x^2-1}+x$ but it turns out too complicated. Could anyone here help me to evaluate the integral? Thanks in advance.

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Hint; Use the Euler substitution; $$x=\frac{u^2+1}{2u}, \mathrm{d}x= \left (1- \frac{u^2+1}{2u^2}\right ) \mathrm{d}u $$ for the integrand, when you substitute back you get $$u=\sqrt{x^2-1}+x$$ The integral becomes;

$$\frac12 \int \frac{u+1}{u} \mathrm{d}u$$

which is pretty easy from now on.

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  • $\begingroup$ You seem to have missed substituting $$dx = \frac{1}{2}(u - u^{-2}) \, du.$$ $\endgroup$ – heropup Sep 11 '14 at 20:32
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The Euler substitution is definitely a smooth approach. But if you are not familiar with that, there is a conventional approach as well. First subtract and add 1 in the numerator so that you can split the fraction. You need to integrate $1$ and the term $\frac{1}{\sqrt{1-x^2}+x-1}$ Now multiply this term top and bottom with its conjugate: $\frac{1}{\sqrt{1-x^2}-(x-1)}$ Things will cancel in the denominator. Verify you end up with $\frac{\sqrt{1-x^2}-x+1}{2(x-1)}$. Now this fraction can be split into three seperate fractional terms. The term with the square root is a simple trig sub (I think you have seen those before) and the other two terms are even easier. This approach is not nearly as quick as the approach of UserX, but I wanted to show that with the ordinary "book"method, it can still be done

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  • $\begingroup$ Wow!? Very simple. Nice observation. +1 $\endgroup$ – Venus Sep 11 '14 at 20:43
  • $\begingroup$ Thanks for posting, I took note of your post and I am going to use your integral in my class pretty soon. $\endgroup$ – imranfat Sep 11 '14 at 20:45
  • $\begingroup$ I am honour. Don't forget to cite my name. :D $\endgroup$ – Venus Sep 11 '14 at 20:47
  • $\begingroup$ A good idea, but I think you've made a few mistakes. The square root is in the reverse order and I don't think the denominator is going to clean up quite as nice. $\endgroup$ – Mike Sep 11 '14 at 22:59
  • $\begingroup$ @Mike My FOIL let me down. I changed that. But it is still integrable though its not so pleasant anymore. Fortunately I have not taken this problem to class and will have to rethink that too. Thanks for the "warning" $\endgroup$ – imranfat Sep 12 '14 at 15:21
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An other solution is to substitute x=sin(u), and then to use the well known relations $cos(u)=\frac{1-t^2}{1+t^2}$ and $sin(u)=\frac{2t}{1+t^2}$ with $t=tan(\frac{u}{2})$ to simplify the problem.

More detailed : $\int\frac{\sqrt{x^2-1}+x}{\sqrt{x^2-1}+x-1}dx = x + \int\frac{1}{\sqrt{x^2-1}+x-1}dx$

$\int^y \frac{1}{\sqrt{x^2-1}+x-1}dx=\int^{arcsin(y)} \frac{cos(x)}{\cos{x}+\sin{x}-1}dx$

With the two relations I gave above, we have $\frac{\cos{x}}{\cos{x}+\sin{x}-1}=1+\frac{\frac{1-t^2}{1+t^2}}{\frac{2t+1-t^2-1-t^2}{1+t^2}}=\frac{1-t^2}{t-t^2}=\frac{1+t}{t}=1+\frac{1}{t}=1+\frac{1}{\tan{\frac{u}{2}}}$

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  • $\begingroup$ Could you please elaborate? Sorry, I can't follow your explanation. $\endgroup$ – Venus Sep 11 '14 at 20:36
  • $\begingroup$ @Venus Pierre is using what's called the Weirestrass substitution, also known as the tangent half angle substitution, another nifty trick that needs some reading on its own. (Note, Weierstrass sub is named after Karl Weierstrass, but the technique was known to others before him as well. Got to cover my bases here....) $\endgroup$ – imranfat Sep 11 '14 at 20:52
  • $\begingroup$ Yes I didn't write to substitute with the tangent half angle because it's a "trick" but the substitution x=sin(u) shall at least seem intuitive in this case... $\endgroup$ – Pierre Alvarez Sep 11 '14 at 20:57
  • $\begingroup$ @imranfat Yes, I get it. +1 Pierre. Thanks for your answer. $\endgroup$ – Venus Sep 12 '14 at 8:57
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Could anyone here help me to evaluate the integral? Thanks in advance.

As a general rule, whenever dealing with expressions containing $\sqrt{x^2\pm1}$, the natural substitution is $x=\cosh t$, $($for $-)$, or $x=\sinh t$, $($for $+)$, since $\cosh^2t-\sinh^2t=1.~$ Then, after using the fact that $\cosh t\pm\sinh t=e^{\large\pm t}$, together with $\cosh't=\sinh t$ and $\sinh't=\cosh t,~$ let $e^t=u$, and employ partial fraction decomposition.

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  • $\begingroup$ Thank you for your answer. You help me again. +1 for sure. $\endgroup$ – Venus Sep 12 '14 at 8:58

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