# 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!

• @Mattos Sir, please could you show me how? – Better World Feb 21 '16 at 11:53
• $u^{2} = a+1$.. – mattos Feb 21 '16 at 11:54

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.

$$\int\frac{1}{a\sqrt{a+1}}\space\text{d}a=$$

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

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

Let $$a = x^2$$

$$\int \frac{2\, dx}{x \sqrt{x^2+1} } = 2 \log \frac{x}{1+ \sqrt{1+x^2}}$$

• It's not any simpler than you started off. – user98186 Feb 21 '16 at 11:57

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.