Help with Integration Problem: \int \frac{\ x^2}{1+x^2} dx

In summary, the conversation was about how to solve the integral \int \frac{x^2}{1+x^2} dx. The options of using integration by parts and partial fractions were discussed, but it was ultimately decided that long division and trigonometric inverses were the best methods to use. The final answer is \int\frac{x^2+1}{x^2+1}-\frac{1}{x^2+1}dx.
  • #1
trap
53
0
can anyone help me with this,

[tex] \int \frac{\ x^2}{1+x^2} dx [/tex]

thanks.

I've tried with integration by parts but does not work..
 
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  • #2
Hmm... I don't quite remember how to use partial fractions for that... but I remember there was a way (at least with a similar problem). You could try

[tex]\int{\frac{x^2}{1+x^2}dx}=\int{\frac{x^2}{(x-i)(x+i)}dx}[/tex]

and use partial fractions from there (but it gets a bit "hairy," as my math teacher always says).
 
  • #3
trap said:
can anyone help me with this,

[tex] \int \frac{\ x^2}{1+x^2} dx [/tex]

thanks.

I've tried with integration by parts but does not work..
Did you try partial fractions?

edit: nevermind read above :)
 
  • #4
trap said:
can anyone help me with this,

[tex] \int \frac{\ x^2}{1+x^2} dx [/tex]

thanks.

I've tried with integration by parts but does not work..

Partial fractions work only for proper fractions. This is not a proper fraction in the sense that the degree of the numerator is the same as the degree of the denominator. So to fix this little problem you need to use long division after which you get.

[tex] 1-\frac{1}{x^2+1} [/tex]

Which is now easy to integrate.

Regards
 
Last edited:
  • #5
Use x^2/(1 + x^2) = (x^2 + 1 - 1)/(1 + x^2) = 1 - 1/(1 + x^2) and integrate that instead...
 
  • #6
In this case partial fractions would lead indeed to a decomposition containing complex numbers.Therefore the method is not good.

Daniel.
 
  • #7
I did this on my calculator and got a simple answer. I think you can do integration by parts. What have you tried?
 
  • #8
Townsend said:
Partial fractions work only for proper fractions. This is not a proper fraction in the sense that the degree of the numerator is the same as the degree of the denominator. So to fix this little problem you need to use long division after which you get.

[tex] 1-\frac{1}{x^2+1} [/tex]

Which is now easy to integrate.

Regards

Integration by parts isn't necessary. Townsend clearly explains the best way to do this.

[tex]\int\frac{x^2}{x^2+1} dx = \int\frac{x^2+1}{x^2+1}-\frac{1}{x^2+1}dx[/tex]

Hint for the second part: trig inverses.
 
  • #9
thank you everyone for the help, i understand now :smile:
 

FAQ: Help with Integration Problem: \int \frac{\ x^2}{1+x^2} dx

What is integration?

Integration is a mathematical process used to find the area under a curve. It involves finding the antiderivative of a function, which is also known as the indefinite integral.

How do you solve an integration problem?

To solve an integration problem, you need to first identify the function and then use a set of rules and techniques to find its antiderivative. These techniques include substitution, integration by parts, and partial fractions.

What is the specific problem of \int \frac{\ x^2}{1+x^2} dx?

The specific problem presented in this integration is known as a rational function. This means that the function is a ratio of two polynomials, in this case, x^2 and 1+x^2. And the goal is to find the antiderivative of this function.

What is the solution to \int \frac{\ x^2}{1+x^2} dx?

The solution to this integration problem is \frac{1}{2}x - \frac{1}{2}tan^{-1}(x) + C, where C is the constant of integration. This can be found using the substitution method by letting u = 1+x^2 and du = 2x dx.

Why is this integration problem useful?

This integration problem may be useful in many areas of science and engineering, such as physics, chemistry, and economics. It can be used to find the total distance traveled, the total volume of a substance, or the total cost of a product, among other applications.

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