Hello, I was wondering if I could get some help with this integral - it would be much appreciated - thanks.
$\displaystyle \int{\frac{1}{\sqrt{x^2-4x+13}}}dx = \int{\frac{1}{\sqrt{(x-2)^2+9}}}dx$
Hello, I was wondering if I could get some help with this integral - it would be much appreciated - thanks.
$\displaystyle \int{\frac{1}{\sqrt{x^2-4x+13}}}dx = \int{\frac{1}{\sqrt{(x-2)^2+9}}}dx$
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!
Hello, slevvio!
$\displaystyle \int \frac{dx}{\sqrt{x^2-4x+13}} \:= \:\int \frac{dx}{\sqrt{(x-2)^2+9}}$
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 . . .
We have a hidden natural logarithm here.
$\displaystyle \int {\frac{{dx}}
{{\sqrt {x^2 - 4x + 13} }}} = \int {\frac{{dx}}
{{\sqrt {(x - 2)^2 + 9} }}} .$
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.