Evaluate indefinite integral $\int \frac{\sqrt{x}+ 1}{\sqrt{x}-1} dx$

Question

How can one evaluate the indefinite integral $$\int \frac{\sqrt{x}+ 1}{\sqrt{x}-1} \,dx$$

First I tried to multiply by $\frac{\sqrt{x}+1}{\sqrt{x}+1}$, but I couldn't get the final result. I also tried to seperate the integral into two $$\int \frac{\sqrt{x}}{\sqrt{x}-1} dx + \int \frac{1}{\sqrt{x}-1} dx$$

which failed to lead to a solution either. Is there another method that can be used?

Answer 1

Hint The form of the denominator suggests the substitution $$t = \sqrt{x} - 1 .$$ In particular, $x$ can be expressed as a polynomial in $t$, so the substitution yields a rational integrand with denominator $t$.

Answer 2

\begin{align} &\int \frac{\sqrt{x}+ 1}{\sqrt{x}-1} \,dx\\ =& \int \frac{x+\sqrt{x}}{\sqrt x(\sqrt{x}-1)} \,dx =\int \frac{(\sqrt{x}+2)(\sqrt x-1)+2}{\sqrt x(\sqrt{x}-1)} \,dx\\ =& \int 1+\frac2{\sqrt x}+\frac{2}{\sqrt x(\sqrt{x}-1)} \,dx =x +4\sqrt x+4\ln|\sqrt x-1|+C \end{align}