- #1
iNCREDiBLE
- 128
- 0
How do I integrate:
\frac{xarctan(x)}{(1+x^2)^2}[/itex]
\frac{xarctan(x)}{(1+x^2)^2}[/itex]
Partial Integration for [tex]\frac{xarctan(x)}{(1+x^2)^2}[/itex]
- Thread starter iNCREDiBLE
- Start date
-
- Tags
- Integration Partial
Click For Summary
SUMMARY
The discussion focuses on the integration of the function [tex]\frac{xarctan(x)}{(1+x^2)^2}[/itex] using integration by parts, often mistakenly referred to as partial integration. Participants emphasize the importance of selecting appropriate functions for u and dv to simplify the integrand. The most effective choice discussed is u = arctan(x) and dv = (x dx) / (1 + x^2)^2, leading to a simpler integrand that can be further integrated. The conversation underscores the necessity of careful analysis before proceeding with integration techniques.
PREREQUISITES- Understanding of integration by parts
- Familiarity with the arctangent function
- Knowledge of basic calculus concepts
- Ability to manipulate rational functions
- Practice integration by parts with various functions
- Explore advanced techniques for integrating rational functions
- Study the properties and applications of the arctangent function
- Learn about common pitfalls in integration and how to avoid them
Students and educators in calculus, mathematicians focusing on integration techniques, and anyone seeking to improve their skills in solving complex integrals.
- #2
quantumdude
Staff Emeritus
Science Advisor
Gold Member
- 5,560
- 24
You have to show us how you started first.
- #3
iNCREDiBLE
- 128
- 0
Tom Mattson said:
How I started first? I tried with partial integration in many different ways..
- #4
quantumdude
Staff Emeritus
Science Advisor
Gold Member
- 5,560
- 24
What do you mean by "partial integration"? I've only heard that term used in reference to integrals of functions of several variables. Did you mean to say, "integration by parts"? If so, then please show us what you did.
- #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 + x2)2, then it's very hard to find v.
If you choose u = x / (1 + x2)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 + x2)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 + x2)2, you can make the integrand look simplier. Now just try it.
You then come up with something like:
\int \frac{dx}{(1 + x ^ 2) ^ 2}, you can again try to integrate it by parts.
Viet Dao,
Now if you choose u = x, and dv = (arctan(x)dx) / (1 + x2)2, then it's very hard to find v.
If you choose u = x / (1 + x2)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 + x2)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 + x2)2, you can make the integrand look simplier. Now just try it.
You then come up with something like:
\int \frac{dx}{(1 + x ^ 2) ^ 2}, you can again try to integrate it by parts.
Viet Dao,
Last edited:
Similar threads
- · Replies 1 ·
- · Replies 7 ·
- · Replies 6 ·
- · Replies 1 ·
- · Replies 3 ·
- · Replies 4 ·
- · Replies 2 ·