Instructions were evaluate the integral.
(x/1-x^4) dx
I know that this form is the tanh inverse 1/1-x^2 but not sure how to substitute and start this problem. Please Help.
To evaluate the integral ∫(x/(1-x^4)) dx, you can notice that the denominator can be factored as (1-x^2)(1+x^2).
Now, you can use partial fraction decomposition to express the integrand as a sum of simpler fractions. The partial fraction decomposition of (x/(1-x^4)) is:
x/(1-x^4) = A/(1-x^2) + B/(1+x^2)
To find the values of A and B, you can multiply both sides of the equation by (1-x^4) and simplify:
x = A(1+x^2) + B(1-x^2)
Now, you can solve for A and B. One way to do this is to equate the coefficients of x^2 on both sides of the equation:
0 = A - B
Next, you can equate the constant terms:
0 = A + B
Solving these equations gives A = B = 0. Therefore, the partial fraction decomposition is:
x/(1-x^4) = 0/(1-x^2) + 0/(1+x^2)
Now, you can integrate each term separately. The integral of 0/(1-x^2) is 0, and the integral of 0/(1+x^2) is also 0.
Therefore, the final result of the integral ∫(x/(1-x^4)) dx is simply 0.
In summary, to evaluate the integral, you need to use partial fraction decomposition to express the integrand as a sum of simpler fractions. However, in this case, the resulting fractions are both 0, so the integral evaluates to 0.