how would I go about integrating:

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

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- Jan 9th 2010, 09:33 PMbhuangCalculus - Integration
how would I go about integrating:

[x^2 / (1+x^2)] dx - Jan 9th 2010, 09:39 PM11rdc11
- Jan 9th 2010, 09:45 PMProve It
- Jan 9th 2010, 10:04 PMbhuang
can you do the long division part?

- Jan 9th 2010, 10:08 PMProve It
- Jan 9th 2010, 10:14 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, 10:16 PM11rdc11
- Jan 9th 2010, 10:19 PMProve It
- Jan 9th 2010, 10:23 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, 10:25 PM11rdc11
- Jan 9th 2010, 10:28 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, 10:34 PM11rdc11
- Jan 9th 2010, 10:36 PMbhuang
Why? What's wrong with doing that?

- Jan 9th 2010, 10:49 PMProve It
Watch carefully:

by definition.

We can also see that

.

This comes from a substitution.

Let so that .

So

.

BUT you are trying to find .

We DO NOT have a formula for that yet because it is not of the form . But you could TRY a substitution.

Watch what happens.

Let .

We can see that .

Can we change into a function that is of the form ? NO, we can not.

So the rule for 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 and seeing if we get . If we do, then .

.

But we know that and we also know that

So

.

So .

Therefore .

No more arguments please! - Jan 9th 2010, 10:54 PM11rdc11