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

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

Step 1 · Rationalize

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

Step 2 · Split

Writing $x^2=(x^2-1)+1$:

$$ \frac{x^2-x}{\sqrt{x^2-1}} = \sqrt{x^2-1} + \frac1{\sqrt{x^2-1}} - \frac x{\sqrt{x^2-1}}. $$

Step 3 · Integrate

Using $\int\sqrt{x^2-1}\,dx=\tfrac x2\sqrt{x^2-1}-\tfrac12\ln|x+\sqrt{x^2-1}|$, $\int dx/\sqrt{x^2-1}=\ln|x+\sqrt{x^2-1}|$, and $\int x\,dx/\sqrt{x^2-1}=\sqrt{x^2-1}$:

$$ I = \frac x2\sqrt{x^2-1} - \frac12\ln\left|x+\sqrt{x^2-1}\right| + \ln\left|x+\sqrt{x^2-1}\right| - \sqrt{x^2-1} + C. $$

$$ I = \frac{x-2}2\sqrt{x^2-1} + \frac12\ln\left|x+\sqrt{x^2-1}\right| + C. $$

Summary Table

Additional Notes

• Major rewrite: the source's substitution $u=\sqrt{(x-1)/(x+1)}$ led to a partial-fractions setup with "detailed calculations omitted for brevity" and arrived at $-\sqrt{x^2-1}-\ln\left|\tfrac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\right|$, which simplifies to $-\sqrt{x^2-1}-\ln|x+\sqrt{x^2-1}|$. Differentiating that gives $-(x+1)/\sqrt{x^2-1}$, which at $x=2$ is $-\sqrt3\approx-1.73$, while the true integrand value is $2\sqrt3/3\approx1.15$ — wrong sign and magnitude. The direct rationalization above avoids the error.
• The domain requires $x>1$ (or $x<-1$).
• The result was verified by differentiation: combining all terms and simplifying gives $\dfrac{x(x-1)}{\sqrt{x^2-1}}$, which equals $x\sqrt{(x-1)/(x+1)}$ exactly.