The integral of $\frac{x+1}{\sqrt {x}-1}$

Question

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

Answer 1

HINT

Write $$\frac{x+1}{\sqrt{x}-1}=\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}-1}+\frac{2}{\sqrt{x}-1}=\sqrt{x}+1+\frac2{\sqrt x-1}$$ and integrate.

Answer 2

If you make the change of variable $$ u=\sqrt{x}-1,\quad x=(u+1)^2, \quad dx=2(u+1)du, $$ then you get an easy rational function to integrate $$ \int\frac{x+1}{\sqrt{x}-1}\:dx=2\int\frac{(u+1)^3+(u+1)}{u}\:du. $$

Answer 3

Another possibility is $x = t^2$ so that $dx = 2t \, dt$

$$\int\frac{2t(t^2+1)}{t-1} dt = \int 2t(t+1) dt +\int 4 dt+\int\frac{4}{t-1} dt$$

I hope you can handle this easily :)

Answer 4

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