242.7.5.88 1/((X+2)sqrt(x^2+4x+3)) complete the square

[karush]

$\large{242.7.5.88}$
$$\displaystyle
I_{88}=\int\frac{dx}{(x+2)\sqrt{{x}^{2}+4x+3}}=
-\arcsin\left(\dfrac{1}{\left|x+2\right|}\right)+C $$
complete the square of
$${x}^{2}+4x+3 ={x}^{2}+4x+3+1-1=(x+2)^2-1 $$
Set $u=(x+2) \ \ du=dx$ then

$$\displaystyle I_{88}=\int\frac{1}{u \sqrt{u^2-1}} \, du \\

u=\cosh(y) \ \ du=\sinh(y) \, dy$$
$$\displaystyle I_{88}
=\int\frac{\sinh(y)}{\cosh(y) \sqrt{cosh^2(y) -1}} \, dy
=\int. ? $$

 

[Greg]

Try the substitution $u=\cosh(y)$ and refer to here for standard integrals of hyperbolic functions and other information.

$$\int\dfrac{1}{(x+2)\sqrt{x^2+4x+3}}\,dx=\tan^{-1}\left(\sqrt{x^2+4x+3}\right)+C$$

is a result.

 

[Greg]

Even simpler, try $u=\sec(y)$.

$$\int\dfrac{1}{(x+2)\sqrt{x^2+4x+3}}\,dx=\cos^{-1}\left(\dfrac{1}{x+2}\right)+C$$

is a result.