MHB 206.8.4.61 int (x^2+2x+4)/(sqrt(x^2-4x)) dx

[karush]

$\tiny{206.8.4.61 \ calculated \ by \ Ti-nspire \ cx \ cas}$
$$I_{61}=\displaystyle
\int\frac{x^2+2x+4}{\sqrt{x^2-4x}} \, dx
=14\ln\left[{\sqrt{{x}^{2}-4x}}+x-2\right]
+\left[\frac{x}{2}+5\right]
\sqrt{{x}^{2}-4x}+C$$
$\text{complete the square}$
$$x^2-4x \implies \left[x-2\right]^2-4$$
$\text{u substitution }$
$$u=x-2 \therefore du=dx \ \ x=u+2 $$
$$I_{61}=\displaystyle
\int\frac{u^2+6u+12}{\sqrt{u^2-4}} \, du$$

 

[Greg]

The solution I have involves hyperbolic functions. I'm going to post it in its entirety so you may familiarise yourself with hyperbolic functions.

$$\int\frac{x^2+2x+4}{\sqrt{x^2-4x}}\,\mathbf dx=\int\frac{x^2+2x+4}{\sqrt{(x-2)^2-4}}\,\mathbf dx$$

$$x-2=2\cosh(z),\quad x=2\cosh(z)+2,\quad z=\arcosh\left(\frac{x-2}{2}\right),$$

$$\mathbf dx=2\sinh(z)\,\mathbf dz$$

$$\int\frac{x^2+2x+4}{\sqrt{(x-2)^2-4}}\,\mathbf dx=\int(2\cosh(z)+2)^2+4\cosh(z)+8\,\mathbf dz$$

$$=\sinh(2z)+12\sinh(z)+14z+C$$

$$=\sqrt{x^2-4x}\left(\frac{x}{2}+5\right)+14\ln\left|x-2+\sqrt{x^2-4x}\right|+C$$

Identities used:

$$\cosh^2(z)-1=\sinh^2(z)$$

$$\cosh^2(z)=\frac{\cosh(2z)+1}{2}$$

$$\sinh(2z)=2\sinh(z)\cosh(z)$$

$$\sinh(\arcosh(z))=\sqrt{z^2-1}$$

$$\arcosh(z)=\ln|z+\sqrt{z^2-1}|$$

Also,

$$\frac{\mathbf d}{\mathbf dz}\sinh(z)=\cosh(z)$$

$$\frac{\mathbf d}{\mathbf dz}\cosh(z)=\sinh(z)$$

Letting Wikipedia do most of the 'work',

hyperbolic function

inverse hyperbolic function

If you have any questions please post. :)

 

[karush]

Thank you, that was very helpful
Saw some other solutions to this but this was far better
You really gave a very concise and comprehensive way to do it

I sent a croped image to a friend and he wanted to know where it came from. Malaho