The integral of $\frac{x+1}{\sqrt {x}-1}$ Can anyone help me with this? I have been racking my brain for days trying to solve this. I tried splitting the equation into two integrals and multiplying by $\frac{\sqrt{x}}{\sqrt{x}}$ for the differential but nothing works. The only hint the book gives me is that $u = \sqrt{x} - 1$.
Thanks for checking.
-Lester

$$\int\frac{x+1}{\sqrt{x}-1}dx=\int\frac{x\sqrt{x}+\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}dx=\int\frac{x\sqrt{x}+\sqrt{x}-2+2}{\sqrt{x}(\sqrt{x}-1)}dx=$$
$$=\int\frac{x\sqrt{x}-x+x-\sqrt{x}+2\sqrt{x}-2}{\sqrt{x}(\sqrt{x}-1)}dx+\int\frac{4}{2\sqrt{x}(\sqrt{x}-1)}dx=$$
$$=\int\frac{x+\sqrt{x}+2}{\sqrt{x}}dx+4\ln|\sqrt{x}-1|=\frac{2}{3}\sqrt{x^3}+x+4\sqrt{x}+4\ln|\sqrt{x}-1|+C.$$
