Indefinite integral $\int \frac{1}{x\sqrt{x+1}}dx$

Question

Is there an elementary solution to the following integration problem:

$$\int \dfrac{1}{x\sqrt{x+1}}dx$$

I tried letting $u=x+1$, but I can't separate it into partial fractions. Any help would be appreciated.

Answer 1

$\textbf{Hint: }$Use substitution $x=u^2-1$, then you have:

$$\int\frac{1}{x\sqrt{x+1}}\;dx=\int\frac{(u^2-1)'}{(u^2-1)\sqrt{u^2-1+1}}\; du=\int \frac{2u}{(u^2-1)|u|}\; du$$

Answer 2

HINT:

Let $\sqrt{x+1}=y\implies\dfrac{dx}{2\sqrt{x+1}}=dy$ and $x+1=y^2\iff x=y^2-1$

$$\frac2{y^2-1}=\frac{y+1-(y-1)}{(y+1)(y-1)}=\cdots$$

Answer 3

Hint

$$u=\sqrt{x+1}$$

and the rest is easy