# Integration

• Dec 26th 2007, 06:02 AM
slevvio
Integration
Hello, I was wondering if I could get some help with this integral - it would be much appreciated - thanks.

$\displaystyle \int{\frac{1}{\sqrt{x^2-4x+13}}}dx = \int{\frac{1}{\sqrt{(x-2)^2+9}}}dx$
• Dec 26th 2007, 06:22 AM
colby2152
Quote:

Originally Posted by slevvio
Hello, I was wondering if I could get some help with this integral - it would be much appreciated - thanks.

$\displaystyle \int{\frac{1}{\sqrt{x^2-4x+13}}}dx = \int{\frac{1}{(x-2)^2+9}}dx$

Do you mean this?

$\displaystyle \int{\frac{1}{\sqrt{x^2-4x+13}}}dx = \int{\frac{1}{\sqrt{(x-2)^2+9}}}dx$

If so, then you can use a trig substitution for x - 2. Set it equal to 3*tan(u). Everything else should fall into place for easy integration!

EDIT: Soroban posted shortly after I did, check out his detailed solution!
• Dec 26th 2007, 07:02 AM
Soroban
Hello, slevvio!

Quote:

$\displaystyle \int \frac{dx}{\sqrt{x^2-4x+13}} \:= \:\int \frac{dx}{\sqrt{(x-2)^2+9}}$

Let: .$\displaystyle x-2 \:=\:3\tan\theta\quad\Rightarrow\quad dx \:=\:3\sec^2\!\theta\,d\theta$

. . and: .$\displaystyle \sqrt{(x-2)^2+9} \;=\;\sqrt{9\tan^2\!\theta + 9} \:=\:\sqrt{9(\tan^2\!\theta + 1)} \:=\:\sqrt{9\sec^2\!\theta} \:=\:3\sec\theta$

Substitute: .$\displaystyle \int\frac{3\sec^2\!\theta\,d\theta}{3\sec\theta} \;=\;\int\sec\theta\,d\theta \;=\;\ln|\sec\theta + \tan\theta| + C$

. . then back-substitute . . .

• Dec 26th 2007, 07:13 AM
Krizalid
We have a hidden natural logarithm here.

Quote:

Originally Posted by slevvio
$\displaystyle \int{\frac{1}{\sqrt{x^2-4x+13}}}dx$

$\displaystyle \int {\frac{{dx}} {{\sqrt {x^2 - 4x + 13} }}} = \int {\frac{{dx}} {{\sqrt {(x - 2)^2 + 9} }}} .$

Substitute $\displaystyle u = x + \sqrt {(x - 2)^2 + 9} - 2 \implies du = 1 + \frac{{x - 2}} {{\sqrt {(x - 2)^2 + 9} }}\,dx.$

So $\displaystyle \frac{{dx}} {{\sqrt {(x - 2)^2 + 9} }} = \frac{{du}} {u}.$

The rest follows.