- #1

- 128

- 0

How do I integrate:

[tex]\frac{xarctan(x)}{(1+x^2)^2}[/itex]

[tex]\frac{xarctan(x)}{(1+x^2)^2}[/itex]

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter iNCREDiBLE
- Start date

- #1

- 128

- 0

How do I integrate:

[tex]\frac{xarctan(x)}{(1+x^2)^2}[/itex]

[tex]\frac{xarctan(x)}{(1+x^2)^2}[/itex]

- #2

Tom Mattson

Staff Emeritus

Science Advisor

Gold Member

- 5,549

- 8

You have to show us how you started first.

- #3

- 128

- 0

Tom Mattson said:You have to show us how you started first.

How I started first? I tried with partial integration in many different ways..

- #4

Tom Mattson

Staff Emeritus

Science Advisor

Gold Member

- 5,549

- 8

- #5

VietDao29

Homework Helper

- 1,424

- 3

Before you integrate something, it's always good to spend some time to look at it closely.

Now if you choose u = x, and dv = (arctan(x)dx) / (1 + x^{2})^{2}, then it's very hard to find v.

If you choose u = x / (1 + x^{2})^{2}, and dv = arctan(x)dx, then you'll get a mess when you try to find du, and obviously, you are complicating the integrand.

And if you choose u = 1 / (1 + x^{2})^{2}, and dv = x arctan(x) dx, then it's hard to find v.

...

And if you choose u = arctan(x), and dv = (x dx) / (1 + x^{2})^{2}, you can make the integrand look simplier. Now just try it.

You then come up with something like:

[tex]\int \frac{dx}{(1 + x ^ 2) ^ 2}[/tex], you can again try to integrate it by parts.

Viet Dao,

Now if you choose u = x, and dv = (arctan(x)dx) / (1 + x

If you choose u = x / (1 + x

And if you choose u = 1 / (1 + x

...

And if you choose u = arctan(x), and dv = (x dx) / (1 + x

You then come up with something like:

[tex]\int \frac{dx}{(1 + x ^ 2) ^ 2}[/tex], you can again try to integrate it by parts.

Viet Dao,

Last edited:

Share: