Integra of x^7/(1-x^4)^2 dx =?

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Answer 1 · 2018-03-21T05:15:33

#1/4[1/(1-x^4)+ln(1-x^4)] + C#

Given: #int x^7/(1-x^4)^2 dx#

Use #u#-substitution:
Let #u = 1-x^4; " " x^4 = 1-u#

#du = -4x^3 dx; " "dx = (du)/(-4x^3)#

#int x^7/(1-x^4)^2 dx = int x^7/u^2 * (du)/(-4x^3) = -1/4 int (x^4 cancel(x^3))/(u^2 cancel(x^3)) du#

#= -1/4 int (1-u)/u^2 du = -1/4 [ int 1/u^2 du - int u/u^2 du]#

#= -1/4[int u^-2 du - int 1/u du]#

# = -1/4 [-u^-1 - ln u] + C#

#= 1/4 [u^-1 + ln u] +C#

#= 1/4 [ 1/(1-x^4) + ln (1-x^4)]#