1. ## Calculus - Integration

how would I go about integrating:

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

2. Originally Posted by bhuang
how would I go about integrating:

[x^2 / (1+x^2)] dx
use long division and get

$\displaystyle \int \bigg(1 - \frac{1}{1+x^2}\bigg)dx$

Then you should remember the second part is arctan

3. Originally Posted by 11rdc11
use long division and get

$\displaystyle \int (1 - \frac{1}{1+x^2})dx$

Then you should remember the second part is arctan
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 $\displaystyle \int{1 - \frac{1}{1 + x^2}\,dx} = x - \arctan{x} + C$.

4. can you do the long division part?

5. Originally Posted by bhuang
can you do the long division part?
Of course you can.

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

6. 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

7. Originally Posted by bhuang
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
No since $\displaystyle u = x^2$ so $\displaystyle du=2xdx$ which you do not have in the numerator.

8. Originally Posted by bhuang
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
It should also be clear that $\displaystyle \arctan{x} \neq \ln{(1 + x^2)}$.

I showed you how to solve it using trigonometric substitution...

9. 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.

10. Originally Posted by bhuang
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.
There isn't another way to solve this but the way Prove It and I showed you. What did you use as your u then?

11. 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).

12. Originally Posted by bhuang
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).
You aren't allowed to do that in math.

13. Why? What's wrong with doing that?

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

15. Originally Posted by bhuang
Why? What's wrong with doing that?

ok so lets say we try to attempt it your way

$\displaystyle \int(1+x^2)^{-1}dx$

$\displaystyle u = (1+x^2)$

$\displaystyle du = 2xdx$

so

$\displaystyle \int u^rdu = \frac{u^{r+1}}{r+1} +c$

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