Integration

**Erghhh** · April 15th 2009, 05:01 PM

Integrate (3+x)/(x^2 -9)^0.5

For now, i would just like to know if i am supposed to use the by parts forumla?

**skeeter** · April 15th 2009, 05:23 PM

Originally Posted by Erghhh

Integrate (3+x)/(x^2 -9)^0.5

For now, i would just like to know if i am supposed to use the by parts forumla?

$\int \frac{x}{\sqrt{x^2-9}} \, dx + \int \frac{3}{\sqrt{x^2-9}} \, dx$

straight up substitution for the first integral , $u = x^2 - 9$

trig sub for the second, $x = 3\sec{t}$

**Soroban** · April 15th 2009, 05:57 PM

Hello, Erghhh!

I'd use a trig substitution . . .

$\int \frac{x+3}{\sqrt{x^2-9}}\,dx$

Let: . $x \:=\:3\sec\theta \quad\Rightarrow\quad dx \:=\:3\sec\theta\tan\theta\,d\theta$
and: . $\sqrt{x^2-9} \:=\:3\tan\theta$

Substitute: . $\int \frac{3\sec\theta + 3}{3\tan\theta}(3\sec\theta\tan\theta\,d\theta) \;=\;3\int(\sec\theta + 1)(\sec\theta\,d\theta)$

. . $=\;3\int(\sec^2\!\theta + \sec\theta)\,d\theta \;=\;3\bigg[\tan\theta + \ln|\sec\theta + \tan\theta| \bigg] + C$

Back-substitute: . $\sec\theta \:=\:\frac{x}{3} \:=\:\frac{hyp}{adj}$
. . Then: . $opp = \sqrt{x^2-9} \quad\Rightarrow\quad \tan\theta \:=\:\frac{\sqrt{x^2-9}}{3}$

We have: . $3\left[\frac{\sqrt{x^2-9}}{3} + \ln\left|\frac{x}{3} + \frac{\sqrt{x^2-9}}{3}\right|\,\right] + C$

. . . . . . . $= \;\sqrt{x^2-9} + 3\ln\left|\frac{x+\sqrt{x^2-9}}{3}\right| + C$

. . . . . . . $= \;\sqrt{x^2-9} + 3\ln\left|x + \sqrt{x^2-9}\right| + C$ .**

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

** .What happened to the denominator, you ask?
. . . . Watch this!

$\sqrt{x^2-9} \:+\: 3\ln\left|\frac{x + \sqrt{x^2-9}}{3}\right| \:+\: C$

. . $= \;\;\sqrt{x^2-9} \:+\: 3\bigg[\ln\left|x + \sqrt{x^2-9}\right| - \ln 3\bigg] \:+\: C$

. . $= \;\;\sqrt{x^2-9} \:+\: 3\ln\left|x + \sqrt{x^2-9}\right| \:-\: \underbrace{3\ln 3 \:+\: C}_{\text{a constant!}}$

. . $= \;\;\sqrt{x^2-9} \:+\:3\ln\left|x + \sqrt{x^2-9}\right| + C$

Textbooks love to pull this on us . . . You've been warned!

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