Solve Integral of 2x/(1-x^4) dx with lxl<1

[kira137]

Homework Statement

Evaluate the following:

Integral of 2x/(1-x^4) dx with lxl<1

The Attempt at a Solution

I tried to use long division to solve the question, then got:

2x/(1-x^4) = (2x^5)/(1-x^4) + 2x

S (2x^5)/(1-x^4) + 2x dx

this is where I seem to be stuck on..
do I use chain rule of letting u = (1-x^4)?

thank you in advance

 

[Bohrok]

Try letting u = x² and then substitute in the original integrand.

 

[Donaldos]

You can also try using the following identity:

$$\frac{2x}{1-x^4}=\frac{1}{2\left(1-x\right)}-\frac{1}{2\left(1+x\right)}+\frac{x}{\left(1+x^2\right)}$$

 

[Mark44]

I tried to use long division to solve the question, then got:
2x/(1-x^4) = (2x^5)/(1-x^4) + 2x

This doesn't make any sense at all. I have no idea how you ended up with that result.

 

[HallsofIvy]

It's called "partial fractions". If you are expected to do this kind of problem, you should have already seen that. [itex]1- x^4= (1- x^2)(1+ x^2)= (1- x)(1+ x)(1+ x^2) so the fraction can be written <br /> $$\frac{2x}{1- x^4}= \frac{A}{1- x}+ \frac{B}{1+ x}+ \frac{Dx+ C}{1+ x^2}$$<br /> There are a number of different wasy of determing the values of A, B, C, and D from that. For example, if you multiply both sides of the equation by $1- x^4$ you get a polynomial equation that must be true for all x. Setting corresponding coefficients equal will give you four equations for A, B, D, and D. More simply, taking x= 1, -1, 0, and, say, 2 will give you four equations.<br /> <br /> However, Bohrok's suggestion of substituting u= x² first is simpler. Once you have it reduced to <br /> $$\int \frac{du}{1- u^2}$$<br /> you can use "partial fractions" to write it as <br /> $$\int \frac{Adu}{1- u}+ \int \frac{Bdu}{1+ u}$$[/itex]

 

[Bohrok]

Or, if you've covered the hyperbolic functions and to make the answer neater, you could use tanh^-1u in getting your answer.

 

[kira137]

Thank all of you for your help.
I haven't learned partial fractions yet so I tried Bohrok's way by letting x^2 = u
This is what I got for answer:

arctan(-x^2) is it right?

I had trouble with solving 1/(1-u^2).. so i just did integral to tan, am I allowed to?

 

[Bohrok]

That's close. ∫1/(1-u²) du = tanh^-1u + C
Then put the integral back in terms of x.