Evaluate the integral: I = integral sqrt((x+1)/(x-1)) dx

Evaluate the integral: $$ I = \int \sqrt{\frac{x+1}{x-1}}\,dx. $$

Step 1 · Rationalize

$$ \sqrt{\frac{x+1}{x-1}} = \sqrt{\frac{(x+1)^2}{(x+1)(x-1)}} = \frac{x+1}{\sqrt{x^2-1}}. $$

Step 2 · Split and integrate

$$ I = \int\frac{x}{\sqrt{x^2-1}}\,dx + \int\frac{dx}{\sqrt{x^2-1}} = \sqrt{x^2-1} + \ln\left|x+\sqrt{x^2-1}\right| + C. $$

Summary Table

Additional Notes

• Major rewrite: the source's approach substituted $u=\sqrt{(x+1)/(x-1)}$ and, after a long partial-fraction and reduction-formula chain, arrived at coefficient $3$ on the logarithm (and offered two "equivalent" boxed forms that are not actually equal to each other, confirming an internal error). Differentiating the source's answer at $x=2$ gives $\approx2.02$, while the true integrand value is $\sqrt3\approx1.73$. The direct rationalization above avoids the error entirely and gives the correct coefficient of $1$.
• The domain requires $x>1$ (or $x<-1$ with sign adjustments).
• The result was verified by differentiation: $\dfrac{d}{dx}\left[\sqrt{x^2-1}+\ln(x+\sqrt{x^2-1})\right] = \dfrac{x}{\sqrt{x^2-1}}+\dfrac1{\sqrt{x^2-1}} = \dfrac{x+1}{\sqrt{x^2-1}} = \sqrt{\dfrac{x+1}{x-1}}$, matching the integrand exactly.