Integration

**slevvio** · December 26th 2007, 06:02 AM

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

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

**colby2152** · December 26th 2007, 06:22 AM

Originally Posted by slevvio

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

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

Do you mean this?

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

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!

**Soroban** · December 26th 2007, 07:02 AM

Hello, slevvio!

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

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

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

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

. . then back-substitute . . .

**Krizalid** · December 26th 2007, 07:13 AM

We have a hidden natural logarithm here.

Originally Posted by slevvio

$\int{\frac{1}{\sqrt{x^2-4x+13}}}dx$

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

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

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

The rest follows.

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