# Calculus Question: $\int\frac{\sqrt{x^2-1}+x}{\sqrt{x^2-1}+x-1}dx$

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.

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.

• You seem to have missed substituting $$dx = \frac{1}{2}(u - u^{-2}) \, du.$$ Sep 11, 2014 at 20:32

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

• Wow!? Very simple. Nice observation. +1 Sep 11, 2014 at 20:43
• Thanks for posting, I took note of your post and I am going to use your integral in my class pretty soon. Sep 11, 2014 at 20:45
• I am honour. Don't forget to cite my name. :D Sep 11, 2014 at 20:47
• 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.
– Mike
Sep 11, 2014 at 22:59
• @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" Sep 12, 2014 at 15:21

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 $\frac{\cos{x}}{\cos{x}+\sin{x}-1}=1+\frac{\frac{1-t^2}{1+t^2}}{\frac{2t+1-t^2-1-t^2}{1+t^2}}=\frac{1-t^2}{t-t^2}=\frac{1+t}{t}=1+\frac{1}{t}=1+\frac{1}{\tan{\frac{u}{2}}}$

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.