How to approach $\int \frac{dx}{\sqrt{x^2+4x+13}}$

Question

My textbook provided solution:

$\int \frac{dx}{\sqrt{\left(x+2\right)^2+9}}$

= $\int \frac{dx}{\sqrt{\left(x+2\right)^2+9}}$

= $\int \frac{ d(x+2)}{\sqrt{\left(x+2\right)^2+9}} $

=$\:\ln \:\left|\left(x+2+\sqrt{\left(x+2\right)^2+9}\right)\right|+C$

Can someone please explain why is that? I don't get it, how can you somehow just get $d(x+2)$?

EDIT: I looked again and it is based on this rule, but I still cannot see how it obtained the answer.

$\int \:\frac{f\:'\left(x\right)dx}{f\left(x\right)}=\int \:\frac{d\:f\:\left(x\right)}{f\left(x\right)}=\ln \left(f\left(x\right)\right)+c$

Answer 1

No. It does not work.

Try the following substitution: $$x+2=3\tan\theta,$$ where $-\frac{\pi}{2}<\theta<\frac{\pi}{2}.$

Answer 2

You can also try this: $$3\sinh(t)=x+2$$ provided you are familiar to hyperbolic function $\sinh(\cdot)$.

Answer 3

hint

If we put $$\frac{e^t-e^{-t}}{2}=u=\sinh(t)$$ then

$$e^t-e^{-t}=2u$$ or

$$(e^t)^2-2ue^t-1=0$$ thus

$$e^t=\frac{2u+\sqrt{4u^2+4}}{2}=u+\sqrt{u^2+1}$$

which gives $$t=\ln(u+\sqrt{u^2+1})$$

Now, as pointed by MRS, replace $ u $ by $\frac{x+2}{3}$.