How do you integrate #int-x/(1-x^2)^4dx# using integration of rational functions by partial fractions?

Question

How do you integrate  #int-x/(1-x^2)^4dx# using integration of rational functions by partial fractions?

Paisley Johnson · Accepted Answer

The answer is #=-1/(6(1-x^2)^3)+C#

There is no need to perform the decomposition into partial fractions.

Perform this integral by substitution

Let #u=1-x^2#, #=>#, #du=-2xdx#

#=>#, #-xdx=1/2du#

Therefore, the integral is

#int(-xdx)/(1-x^2)^4=1/2int(du)/u^4#

#=1/2(-1/(3u^3))#

#=-1/(6(1-x^2)^3)+C#

Emilia Aguilar · Answer

undefined To integrate $\int \frac{-x}{(1-x^2)^4} \, dx$ using partial fractions, follow these steps:

1. Express the integrand as a sum of partial fractions.
2. Find the partial fraction decomposition by setting $\frac{-x}{(1-x^2)^4}$ equal to $\frac{A}{1-x^2} + \frac{Bx + C}{(1-x^2)^2} + \frac{Dx + E}{(1-x^2)^3} + \frac{Fx + G}{(1-x^2)^4}$.
3. Clear the fractions by multiplying both sides by $(1-x^2)^4$.
4. Equate coefficients of like terms to solve for $A$, $B$, $C$, $D$, $E$, $F$, and $G$.
5. Once you find the values of $A$, $B$, $C$, $D$, $E$, $F$, and $G$, rewrite the original integral as a sum of integrals of simpler fractions.
6. Integrate each term separately.
7. Combine the integrals to get the solution.

The final answer should be in the form of a sum of simpler integrals.