Evaluate $\int\frac{da}{a\sqrt{a+1}}$ 
$$\int\dfrac{da}{a\sqrt{a+1}}$$

I don't know how to solve this integral. The fact that $\dfrac1a$ is the derivative of $\ln(a)$ and $\dfrac{1}{\sqrt{a+1}}$ is the derivative of $\cos^{-1}a$ suggested Integration by Parts.
$$\int\dfrac{da}{a\sqrt{a+1}} = \dfrac{\ln(a)}{\sqrt{a+1}}-\int $$
However I got stuck after this.
Any help with the Integral would be greatly appreciated. Many thanks!
 A: Substitute:
$$x^2=a+1\implies 2x\,dx=da\implies$$
$$\int\frac{da}{a\sqrt{a+1}}=2\int\frac{x\,dx}{(x^2-1)x}=2\int\frac{dx}{(x-1)(x+1)}$$
and now you can do simple fractions.
A: Let  $a = x^2$
$$ \int \frac{2\, dx}{x \sqrt{x^2+1} } = 2 \log \frac{x}{1+ \sqrt{1+x^2}}$$
A: $$\int\frac{1}{a\sqrt{a+1}}\space\text{d}a=$$

Substitute $u=a+1$ and $\text{d}u=\text{d}a$:

$$\int\frac{1}{(u-1)\sqrt{u}}\space\text{d}u=$$

Substitute $s=\sqrt{u}$ and $\text{d}s=\frac{1}{2\sqrt{u}}\space\text{d}u$:

$$2\int\frac{1}{s^2-1}\space\text{d}s=-2\int\frac{1}{1-s^2}\space\text{d}s=-2\text{arctanh}\left(s\right)+\text{C}=$$
$$-2\text{arctanh}\left(\sqrt{u}\right)+\text{C}=-2\text{arctanh}\left(\sqrt{a+1}\right)+\text{C}$$
A: Trig sub approach:
$$\int\frac{1}{a\sqrt{a+1}}\space\text{d}a=$$

Substitute $a=\tan{\theta}$ and $da=\frac{2\tan{\theta}}{\cos^2{\theta}}du $:

$$\int\frac{1}{\tan^2{\theta}\sqrt{\tan^2{\theta}+1}}\frac{2\tan{\theta}}{\cos^2{\theta}} d\theta =$$
$$=\int\frac{1}{\tan^2{\theta}\frac{1}{\cos{\theta}}}\frac{2\tan{\theta}}{\cos^2{\theta}} d\theta=$$
$$=2\int \frac{1}{\tan{\theta}\cos{\theta}}d\theta=$$
$$= 2\int \frac{1}{\sin{\theta}}d\theta$$

Now with these integral, we do:
$$\int \frac{1}{\sin{\theta}}d\theta = \int \frac{\sin{\theta}}{\sin^2{\theta}}d\theta = \int \frac{{\sin{\theta}}}{1-\cos^2{\theta}}d\theta$$

Using substitution $t = \cos{\theta}$ and $dt$ = $-\sin{\theta} d\theta$:

$$-\int \frac{1}{1-t^2}dt = \int \frac{-1}{(1-t)(1+t)}dt = \int \frac{1}{2(t-1)} - \frac{1}{2(1+t)} dt = \frac{1}{2} \ln{(t-1)} - \frac{1}{2} \ln{(t+1)} - C$$(-C because i'm special)

So the original integral is:
$$\int \frac{1}{a\sqrt{a+1}} = \int \ln{(t-1)} - \ln{(t-1)}dt = \ln{\Bigl(\frac{\arctan{(\cos{a})} -1}{\arctan{(\cos{a})}+1} \Bigr)} - C $$
Not really a nice solution, but still.
A: Note
$$\int\dfrac{da}{a\sqrt{a+1}}\overset{a=\frac1{t^2}}=-\int \frac2{\sqrt{1+t^2}}dt=-2\sinh^{-1}t+C
$$
A: $$
\begin{aligned}
\int \frac{d a}{a \sqrt{a+1}}=& 2 \int \frac{d(\sqrt{a+1})}{a}\\
=& 2 \int \frac{d \sqrt{a+1}}{(\sqrt{a+1})^{2}-1} \\
=& \ln \left|\frac{\sqrt{a+1}-1}{\sqrt{a+1}+1}\right|+C
\end{aligned}
$$
