how would I go about integrating:

[x^2 / (1+x^2)] dx

Printable View

- Jan 9th 2010, 08:33 PMbhuangCalculus - Integration
how would I go about integrating:

[x^2 / (1+x^2)] dx - Jan 9th 2010, 08:39 PM11rdc11
- Jan 9th 2010, 08:45 PMProve It
And if you didn't know that, you could use trigonometric substitution.

$\displaystyle \int{1 - \frac{1}{1 + x^2}\,dx} = \int{1\,dx} - \int{\frac{1}{1 + x^2}\,dx}$

$\displaystyle = x - \int{\frac{1}{1 + x^2}\,dx}$.

To find $\displaystyle \int{\frac{1}{1 + x^2}\,dx}$

Let $\displaystyle x = \tan{\theta}$ so that $\displaystyle dx = \sec^2{\theta}\,d\theta$.

Also note that $\displaystyle \theta = \arctan{x}$.

So the integral becomes

$\displaystyle \int{\frac{1}{1 + \tan^2{\theta}}\,\sec^2{\theta}\,d\theta}$

$\displaystyle = \int{\frac{1}{\sec^2{\theta}}\,\sec^2{\theta}\,d\t heta}$

$\displaystyle = \int{1\,d\theta}$

$\displaystyle = \theta + C$

$\displaystyle = \arctan{x} + C$.

So $\displaystyle \int{1 - \frac{1}{1 + x^2}\,dx} = x - \arctan{x} + C$. - Jan 9th 2010, 09:04 PMbhuang
can you do the long division part?

- Jan 9th 2010, 09:08 PMProve It
- Jan 9th 2010, 09:14 PMbhuang
i haven't learned arctan or trigonometry for integration. so can I not simply integrate 1/(1+x^2) as ln(1+x^2)?

so my final answer would be

x-ln(1+x^2)+c - Jan 9th 2010, 09:16 PM11rdc11
- Jan 9th 2010, 09:19 PMProve It
- Jan 9th 2010, 09:23 PMbhuang
I don't want to use something I haven't learned, it will confuse me before my test. I might not even remember it if I come across a similar question. I never made u=x^2.

- Jan 9th 2010, 09:25 PM11rdc11
- Jan 9th 2010, 09:28 PMbhuang
I didn't have a u. I went from 1-1/(1+x^2) and I integrated.

1 becomes x, when integrated. And 1/(1+x^2) I made (1+x^2)^-1 so I could integrate that into ln(1+x^2). - Jan 9th 2010, 09:34 PM11rdc11
- Jan 9th 2010, 09:36 PMbhuang
Why? What's wrong with doing that?

- Jan 9th 2010, 09:49 PMProve It
Watch carefully:

$\displaystyle \int{\frac{1}{x}\,dx} = \ln{|x|} + C$ by definition.

We can also see that

$\displaystyle \int{\frac{1}{ax + b}\,dx} = \frac{1}{a}\ln{|ax + b|}$.

This comes from a $\displaystyle u$ substitution.

Let $\displaystyle u = ax + b$ so that $\displaystyle \frac{du}{dx} = a$.

So $\displaystyle \int{\frac{1}{ax + b}\,dx} = \frac{1}{a}\int{\frac{a}{ax + b}\,dx}$

$\displaystyle = \frac{1}{a}\int{\frac{1}{u}\,\frac{du}{dx}\,dx}$

$\displaystyle = \frac{1}{a}\int{\frac{1}{u}\,du}$

$\displaystyle = \frac{1}{a}\ln{|u|} + C$

$\displaystyle = \frac{1}{a}\ln{|ax + b|} + C$.

BUT you are trying to find $\displaystyle \int{\frac{1}{1 + x^2}\,dx}$.

We DO NOT have a formula for that yet because it is not of the form $\displaystyle \frac{1}{ax + b}$. But you could TRY a $\displaystyle u$ substitution.

Watch what happens.

Let $\displaystyle u = 1 + x^2$.

We can see that $\displaystyle \frac{du}{dx} = 2x$.

Can we change $\displaystyle \int{\frac{1}{1 + x^2}\,dx}$ into a function that is of the form $\displaystyle \int{\frac{2x}{1 + x^2}\,dx}$? NO, we can not.

So the rule for $\displaystyle \ln$ DOES NOT WORK in this case.

Now see my post regarding trigonometric substitution to show you the method that DOES work.

Alternatively, let's try differentiating $\displaystyle \arctan{x}$ and seeing if we get $\displaystyle \frac{1}{1 + x^2}$. If we do, then $\displaystyle \int{\frac{1}{1 + x^2}\,dx} = \arctan{x}$.

$\displaystyle y = \arctan{x}$

$\displaystyle \tan{y} = x$

$\displaystyle \frac{d}{dx}(\tan{y}) = \frac{d}{dx}(x)$

$\displaystyle \frac{dy}{dx}\,\frac{d}{dy}(\tan{y}) = 1$

$\displaystyle \sec^2{y}\,\frac{dy}{dx} = 1$

$\displaystyle \frac{dy}{dx} = \cos^2{y}$.

But we know that $\displaystyle y = \arctan{x}$ and we also know that $\displaystyle \cos{\theta} = \frac{1}{\sqrt{1 + \tan^2{\theta}}}$

So $\displaystyle \frac{dy}{dx} = \left(\frac{1}{\sqrt{1 + \tan^2{y}}}\right)^2$

$\displaystyle = \frac{1}{1 + \tan^2{y}}$

$\displaystyle = \frac{1}{1 + (\tan{\arctan{x}})^2}$

$\displaystyle = \frac{1}{1 + x^2}$.

So $\displaystyle \frac{d}{dx}(\arctan{x}) = \frac{1}{1 + x^2}$.

Therefore $\displaystyle \int{\frac{1}{1 + x^2}\,dx} = \arctan{x} + C$.

No more arguments please! - Jan 9th 2010, 09:54 PM11rdc11