Integrating $\int \frac{1}{\sqrt{x}(x+1)}\mathrm{d}x$

Question

$$\int \frac{1}{\sqrt{x}(x+1)}\mathrm{d}x$$

I've tried substituting $u = \sqrt{x}$ giving $\mathrm{d}u = \frac{1}{2u}$ hence the integral becomes

$$2 \cdot\int \frac{1}{u^2(u^2+1)}\mathrm{d}u,$$ but after this I got stuck.

Any hints or help would be appreciated.

Regards Aditya

Answer 1

Another approach: Substitute $x = u^2$ to obtain

$$\int \frac{1}{\sqrt{x}(x+1)}\mathrm{d}x=\int \frac{1}{u(u^2+1)}\mathrm{d}x. $$

Now $\mathrm{d}x = 2u\cdot \mathrm{d}u$, and therefore

$$\int \frac{2}{ u^2+1 }\mathrm{d}u.$$ Let $u = \tan(a)$, then $\mathrm{d}u =\sec^2(a)\mathrm{d}a$. Thus,

$$\int \frac{2}{ \sec^2(a) }\sec^2(a)\mathrm{d}a=2\int \mathrm{d}a=2a+c.$$

Since $a = \arctan(u)$ and $u = \sqrt{x}$, the final answer is

$$\int \frac{1}{\sqrt{x}(x+1)}\mathrm{d}x=2 \arctan(\sqrt{x})+c.$$

Answer 2

$$ \begin{aligned} \int \frac{1}{\sqrt{x}(x+1)} d x &=2 \int \frac{1}{x+1} d (\sqrt{x}) \\ &=2 \int \frac{1}{(\sqrt{x})^{2}+1} d(\sqrt{x}) \\ &=2 \tan ^{-1}( \sqrt{x})+C \end{aligned} $$