How Do You Integrate \( \frac{x}{(1-x^2)} \) dx?

It should be \sqrt{y} instead of \sqrt{x}.In summary, the conversation discusses how to solve the integral $\int \frac{x}{(1-x^2)} dx$ using integration by parts and substitution methods. It is suggested to substitute $y = x^2$ and then use the formula $\int \frac{1}{2}\frac{1}{1-y}dy$ to make the integral more manageable. Another suggestion is to let $u = 1-x^2$ and use the formula $\int \frac{1}{u} du$ to solve the integral. The conversation concludes with the realization that the first response had a typo and the correct formula should be $\int \frac{1}{
  • #1
tomwilliam2
117
2

Homework Statement


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

Homework Equations



Integration by parts, by substitution, etc.

The Attempt at a Solution



I just can't remember how to begin this integration. I tried doing integration by parts, where
$$a(x) = x$$
$$a'(x) = 1$$
$$b(x) = $$
$$b'(x) = \frac{1}{(1-x^2)}$$
$$\int \frac{x}{(1-x^2)}dx = xb(x) - \int \frac{1}{(1-x^2)} dx$$

But couldn't work out what form b(x) should take.
Thanks in advance for any help
 
Physics news on Phys.org
  • #2
If you have seen enough of these integrals, they can essentially done "by observation"

If not, substitution is a good method to use. Substitute [itex]y = x^{2}[/itex]. Then [itex]dy = 2x dx[/itex]
Then the integral becomes:
[tex]\int \frac{1}{2}\frac{1}{1-y}dy[/tex]

Does that now look more doable?
 
Last edited:
  • #3
Why not let u= 1- x^2 itself? Then du= -2xdx so that xdx= -(1/2)du do that
[itex]-\frac{1}{2}\int \frac{du}{u}[/itex]
 
  • #4
Ah yes, thanks...that makes sense. I wasn't getting where the sqrt cam from in the first response.
Thanks to both of you
 
  • #5
tomwilliam2 said:
I wasn't getting where the sqrt cam from in the first response
Oops, typo, my bad.
 

FAQ: How Do You Integrate \( \frac{x}{(1-x^2)} \) dx?

What is integration by parts?

Integration by parts is a method used in calculus to solve integrals that involve products of functions. It is based on the product rule for differentiation, which states that the derivative of a product of two functions is equal to the first function times the derivative of the second function plus the second function times the derivative of the first function.

When do we use integration by parts?

Integration by parts is used when the integral involves a product of functions that cannot be easily solved using other methods such as substitution or trigonometric identities. It is especially useful for integrals involving logarithmic, exponential, or polynomial functions.

How do you solve an integral using integration by parts?

To solve an integral using integration by parts, you must first identify which function can be differentiated and which function can be integrated. Then, you can use the formula: ∫ u dv = uv - ∫ v du, where u is the function that can be differentiated and dv is the function that can be integrated. You will then continue to integrate the remaining integral until it can be solved, or until the integral repeats itself.

Can integration by parts be used for definite integrals?

Yes, integration by parts can be used for both indefinite and definite integrals. When solving a definite integral using integration by parts, you will need to evaluate the limits of integration and substitute them into the final solution.

Are there any tips for making integration by parts easier?

One helpful tip for making integration by parts easier is to choose u and dv in a way that will make the resulting integral simpler to solve. This can involve choosing u to be the most complicated function in the integral or choosing dv to be a function that becomes simpler when differentiated.

Similar threads

Replies
6
Views
491
Replies
22
Views
2K
Replies
54
Views
10K
Replies
5
Views
2K
Replies
4
Views
1K
Replies
12
Views
1K
Back
Top