# The integral of 1/ 11+x^2

• May 5th 2009, 04:12 PM
Fallen186
The integral of 1/ 11+x^2
Problem:
$\displaystyle \int\frac{1}{11+x^2}dx$

Attempt:
Quote:

$\displaystyle x = \sqrt{11}*tan(\theta)$
$\displaystyle dx = \frac{\sqrt{11} *d\theta}{cos^2\theta}$
$\displaystyle \int\frac{1}{11+(\sqrt{11}+tan(\theta))^2} * \frac{\sqrt{11} *d\theta}{cos^2\theta}$

$\displaystyle \int\frac{\sqrt{11}*d\theta}{(11+\sqrt{11}*tan(\th eta))^2*cos^2(\theta)}$

$\displaystyle \int\frac{\sqrt{11}*d\theta}{(11+11*tan^2(\theta)) *cos^2(\theta)}$

$\displaystyle \int\frac{\sqrt{11}*d\theta}{11(1+\frac{sin^2(\the ta)}{cos^2(\theta})*cos^2(\theta)}$

$\displaystyle \int\frac{\sqrt{11}*d\theta}{11*(cos^2(\theta)+sin ^2(\theta))}$

$\displaystyle \int\frac{\sqrt{11}*d\theta}{11}$
$\displaystyle \frac{\sqrt{11}*\theta}{11} + C$

$\displaystyle \frac{\sqrt{11}}{11}*arctan(\frac{x}{\sqrt{11}})+C$

$\displaystyle \frac{1}{\sqrt{11}}*arctan(\frac{x}{\sqrt{11}})+C$

Can someone help?
• May 5th 2009, 04:25 PM
Jameson
I don't see how the second line easily follows:

$\displaystyle \int\frac{\sqrt{11}*d\theta}{(11+11*tan^2(\theta)) *cos^2(\theta)}$

Did you square the sum in the denominator? Maybe you just did a lot of simplification but I don't see how you got to the this line.

edit: Well don't listen to me, skeeter says you're right, so I believe him. I didn't think you were necessarily wrong, I just didn't see the work between lines.
• May 5th 2009, 04:27 PM
skeeter
you are correct ... the "correct" answer is not.
• May 5th 2009, 04:58 PM
Fallen186
Quote:

Originally Posted by Jameson
I don't see how the second line easily follows:

$\displaystyle \int\frac{\sqrt{11}*d\theta}{(11+11*tan^2(\theta)) *cos^2(\theta)}$

Did you square the sum in the denominator? Maybe you just did a lot of simplification but I don't see how you got to the this line.

edit: Well don't listen to me, skeeter says you're right, so I believe him. I didn't think you were necessarily wrong, I just didn't see the work between lines.

I simplified the dx into the big fraction and I foiled the $\displaystyle (\sqrt{11}*tan(\theta))^2$
• May 5th 2009, 05:18 PM
Pinkk
Uhh...$\displaystyle \frac{1}{\sqrt{11}}=\frac{\sqrt{11}}{11}$. Those two answers are equal to one another.

And I don't know if this has already been defined in your class, but:

$\displaystyle \int \frac{du}{a^{2}+u^{2}}=\frac{1}{a}tan^{-1}(\frac{u}{a})+C$

And for this problem, $\displaystyle a=\sqrt{11},u=x$.