Integration Challenge

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June 18th 2006, 12:31 AM
nath_quam

Integration Challenge

Find the integral of (x^2+3)/(x^2+9)^1/2
June 18th 2006, 06:53 AM
malaygoel

Quote:

Originally Posted by nath_quam

Find the integral of (x^2+3)/(x^2+9)^1/2

$\int \frac {x^2+3}{\sqrt{x^2+9}}dx$
Change $(x^2+3)$ into $(x^2 +9) - 6$ and you will get the answer.
June 18th 2006, 02:42 PM
Soroban

Hello, nath_quam!

Anyway we do it, it's not simple . . .

Quote:

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

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

Substitute: $\int\frac{9\tan^2\!\theta + 3}{3\sec\theta}\,3\sec^2\!\theta\,d\theta \;= \;\int(9\tan^2\!\theta + 3)\sec\theta\,d\theta$

. . $= \;\int[9(\sec^2\!\theta - 1) + 3]\sec\theta\,d\theta \;=\;\int(9\sec^2\!\theta - 6)\sec\theta\,d\theta<br />$

. . $=\;9\int\sec^3\!\theta\,d\theta - 6\int\sec\theta\,d\theta$

The first integral must be done by parts (or a formula can be applied);

. . $\frac{9}{2}\bigg(\sec\theta\tan\theta + \ln|\sec\theta + \tan\theta|\bigg) - 6\ln\left|\sec\theta + \tan\theta\right| + C$

. . $= \;\frac{9}{2}\sec\theta\tan\theta - \frac{3}{2}\ln\left|\sec\theta + \tan\theta\right| + C$

. . $= \;\frac{3}{2}\bigg(3\sec\theta\tan\theta - \ln\left|\sec\theta + \tan\theta\right|\bigg) + C<br />$
June 19th 2006, 04:21 AM
malaygoel

Hello Soroban!
How are you.

Quote:

Originally Posted by Soroban

Hello, nath_quam!

Anyway we do it, it's not simple . . .

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

Substitute: $\int\frac{9\tan^2\!\theta + 3}{3\sec\theta}\,3\sec^2\!\theta\,d\theta \;= \;\int(9\tan^2\!\theta + 3)\sec\theta\,d\theta$

. . $= \;\int[9(\sec^2\!\theta - 1) + 3]\sec\theta\,d\theta \;=\;\int(9\sec^2\!\theta - 6)\sec\theta\,d\theta<br />$

. . $=\;9\int\sec^3\!\theta\,d\theta - 6\int\sec\theta\,d\theta$

The first integral must be done by parts (or a formula can be applied);

. . $\frac{9}{2}\bigg(\sec\theta\tan\theta + \ln|\sec\theta + \tan\theta|\bigg) - 6\ln\left|\sec\theta + \tan\theta\right| + C$

. . $= \;\frac{9}{2}\sec\theta\tan\theta - \frac{3}{2}\ln\left|\sec\theta + \tan\theta\right| + C$

. . $= \;\frac{3}{2}\bigg(3\sec\theta\tan\theta - \ln\left|\sec\theta + \tan\theta\right|\bigg) + C<br />$

Please give a final touch to your post(change theta into x in the answer you got).
Keep Smiling
Malay
June 19th 2006, 05:38 AM
Soroban

Hello, Malay!

Quote:

Please give a final touch to your post (change theta into x in the answer you got).

I'd be happy to . . . I'll simplify it to the bitter, bitter end.

We have: . $\frac{3}{2}\left(3\sec\theta\tan\theta - \ln \left|\sec\theta + \tan\theta\right|\right) + C$

From $x = 3\tan\theta$ , we have: . $\tan\theta = \frac{x}{3}\quad\Rightarrow\quad\sec\theta = \frac{\sqrt{x^2+9}}{3}$

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

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

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

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

. . $= \;\frac{1}{2}x\sqrt{x^2+9} \:- \:\frac{3}{2}\ln\left|x + \sqrt{x^2+9}\right| \,+ \,\underbrace{\frac{3}{2}\ln 3 \,+ \,C}$
. . . . . . . . . . . . . . . . . . . . . . . . . $\text{This is a constant}$

Therefore: . $\frac{1}{2}x\sqrt{x^2+9} \:- \:\frac{3}{2}\ln\left|x + \sqrt{x^2+9}\right| \,+ \,C$

See what I mean?