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How would one go about solving

$$\int\frac{1}{\sqrt{x}-1}dx$$

I tried two ways.

A) Multiply denominator and numerator by the conjugate of the denominator, then treat the difference of squares.

$$\int\frac{1}{\sqrt{x}-1}dx=\int\frac{\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+1)}dx=\int\frac{\sqrt{x}+1}{x-1}dx=\int\frac{\sqrt{x}}{x-1}+\int\frac{1}{x-1}$$

While the second integral of the resulting expression is very simple, I find no substitution to help me deal with $\int\frac{\sqrt{x}}{x-1}$.

B) Stuck in way A, I decided to try another approach. So I went back to the initial integral and let $u=\sqrt{x}-1$, so $du=\frac{dx}{2\sqrt{x}}$. I found no way to make $du$ appear in my expression. Let's see making $\frac{1}{2}$ "appear".

$$\int\frac{1}{\sqrt{x}-1}dx=2\int\frac{1}{2\sqrt{x}-2}dx$$

Still $du$ does not appear by itself, but is just a term in the denominator, and I can't seem to figure out how to isolate it while not changing the actual value of the expression.

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Substitute $\sqrt{x}-1 = t$ so $x = (t+1)^2$ and $dx = 2(t+1)\;dt$. Then $$ \int\frac{dx}{\sqrt{x}-1} = \int\frac{2(t+1)\;dt}{t} = \int\left(2+\frac{2}{t}\right) dt $$ In general, to get rid of something complicated (for example, in the denominator, or inside a square-root): use that as your new variable.

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  • $\begingroup$ Thanks, GEdgar. Your answer is the same in result as that of Yves, to whom I'm also grateful, but you were a little bit more detailed and 'step by step' on it, so I'll mark you as the best answer. $\endgroup$ – lafinur Jan 10 at 22:58
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With $x=t^2$,

$$\int\frac{dx}{\sqrt x-1}=\int\frac{2t\,dt}{t-1}=2\int dt+2\int\frac{dt}{t-1}.$$


And by your proposal $x=(u+1)^2$,

$$\int\frac{dx}{\sqrt x-1}=\int\frac{2(u+1)\,du}{u}=2\int du+2\int\frac{du}{u}.$$

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If you have any square roots and 'normal'degrees , take the substitution equal to the degree of the root , if there are multiple roots take the NZS of the roots. Works most of the time, as in this case, taking $x=t^2$ helped. I wont repeat the work of the collegue above.

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