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

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

Explanation of Method

Substituting $u=\sqrt x$ clears the square root, and polynomial long division reduces the resulting improper rational function to a polynomial plus a simple log term.

Substitute $u=\sqrt x$.
Divide $u(u+1)$ by $(u-1)$.
Integrate term by term.
Back-substitute.

Step 1 · Substitute

Let $u=\sqrt x$, so $x=u^2$ and $dx=2u\,du$:

$$ I = \int\frac{u+1}{u-1}\cdot2u\,du = 2\int\frac{u^2+u}{u-1}\,du. $$

Step 2 · Divide

$$ \frac{u^2+u}{u-1} = u+2+\frac2{u-1}. $$

Step 3 · Integrate and back-substitute

$$ 2\int\left(u+2+\frac2{u-1}\right)du = u^2+4u+4\ln|u-1|+C. $$

$$ I = x + 4\sqrt x + 4\ln\left|\sqrt x-1\right| + C. $$

Summary Table

Piece	Antiderivative
$u+2$	$u^2/2+2u$
$2/(u-1)$	$2\ln\|u-1\|$

Additional Notes

The domain requires $x\geq0,\ x\neq1$.
The result was verified by differentiation: $\dfrac{d}{dx}\left[x+4\sqrt x+4\ln|\sqrt x-1|\right] = 1+\dfrac2{\sqrt x}+\dfrac2{\sqrt x(\sqrt x-1)} = 1+\dfrac2{\sqrt x-1} = \dfrac{\sqrt x+1}{\sqrt x-1}$, matching the integrand.

Result

The solution to the integral is:

$$ I = x + 4\sqrt x + 4\ln\left|\sqrt x-1\right| + C. $$