Let 3 /( (1 + x) * (1 - 2x))3(1+x)⋅(1−2x) be = (A /(1 + x) + B / (1 - 2x))(A1+x+B1−2x)
Expanding the Right hand side, we get
(A * (1 - 2x) + B * (1 + x)) / ((1 + x) * (1 - 2x)A⋅(1−2x)+B⋅(1+x)(1+x)⋅(1−2x)
Equating, we get
(A * (1 - 2x) + B * (1 + x)) / ((1 + x) * (1 - 2x)A⋅(1−2x)+B⋅(1+x)(1+x)⋅(1−2x) = 3 /( (1 + x) * (1 - 2x))3(1+x)⋅(1−2x)
ie A * (1 - 2x) + B * (1 + x) = 3A⋅(1−2x)+B⋅(1+x)=3
or A - 2Ax + B + Bx = 3A−2Ax+B+Bx=3
or (A + B) + x*(-2A + B) = 3(A+B)+x⋅(−2A+B)=3
equating the coefficient of x to 0 and equating constants, we get
A + B = 3A+B=3 and
-2A + B = 0−2A+B=0
Solving for A & B, we get
A = 1 and B = 2A=1andB=2
Substituting in the integration, we get
int 3 /( (1 + x) * (1 - 2x))dx∫3(1+x)⋅(1−2x)dx = int (1 /(1 + x) + 2 / (1 - 2x))
dx∫(11+x+21−2x)dx
= int(1 / (1 + x)) dx + int(2 / (1 - 2x))dx∫(11+x)dx+∫(21−2x)dx
= ln(1 + x) + 2 * ln(1 - 2x) * (-1 / 2)ln(1+x)+2⋅ln(1−2x)⋅(−12)
= ln(1 + x) - ln(1 - 2x)ln(1+x)−ln(1−2x)
= ln((1 + x) /(1 - 2x)) + Cln(1+x1−2x)+C
