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!

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.
