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

Question

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

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!

Answer 1

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.

Answer 2

Let $a = x^2$

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

Answer 3

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

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

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

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

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

Answer 4

Note $$\int\dfrac{da}{a\sqrt{a+1}}\overset{a=\frac1{t^2}}=-\int \frac2{\sqrt{1+t^2}}dt=-2\sinh^{-1}t+C $$

Answer 5

$$ \begin{aligned} \int \frac{d a}{a \sqrt{a+1}}=& 2 \int \frac{d(\sqrt{a+1})}{a}\\ =& 2 \int \frac{d \sqrt{a+1}}{(\sqrt{a+1})^{2}-1} \\ =& \ln \left|\frac{\sqrt{a+1}-1}{\sqrt{a+1}+1}\right|+C \end{aligned} $$