Let #3 /( (1 + x) * (1 - 2x))# be = #(A /(1 + x) + B / (1 - 2x))#
Expanding the Right hand side, we get
#(A * (1 - 2x) + B * (1 + x)) / ((1 + x) * (1 - 2x)#
Equating, we get
#(A * (1 - 2x) + B * (1 + x)) / ((1 + x) * (1 - 2x)# = #3 /( (1 + x) * (1 - 2x))#
ie #A * (1 - 2x) + B * (1 + x) = 3#
or #A - 2Ax + B + Bx = 3#
or #(A + B) + x*(-2A + B) = 3#
equating the coefficient of x to 0 and equating constants, we get
#A + B = 3# and
#-2A + B = 0#
Solving for A & B, we get
#A = 1 and B = 2#
Substituting in the integration, we get
#int 3 /( (1 + x) * (1 - 2x))dx# = #int (1 /(1 + x) + 2 / (1 - 2x))
dx#
= #int(1 / (1 + x)) dx + int(2 / (1 - 2x))dx#
= #ln(1 + x) + 2 * ln(1 - 2x) * (-1 / 2)#
= #ln(1 + x) - ln(1 - 2x)#
= #ln((1 + x) /(1 - 2x)) + C#
