To integrate, we should first convert #1/((1+x)(1-2x))# in to partial fractions. Let
#1/((1+x)(1-2x))hArrA/(1+x)+B/(1-2x)#. Simplifying RHS
=#[A(1-2x)+B(1+x)]/((1+x)(1-2x)# or
=#[A-2Ax+B+Bx]/((1+x)(1-2x)#
=#[(B-2A)x+(A+B)]/((x+3)(x-7)(x+4))#
Hence #B-2A=0#, #A+B=1# or #A=1-B#
Hence #B-2(1-B)=0# or #3B=2# or
#B=2/3# and hence #A=1/3#
Hence #1/((1+x)(1-2x))hArr1/(3(1+x))+2/(3(1-2x))#
Hence #int1/((1+x)(1-2x))dx# =
#int[1/(3(1+x))+2/(3(1-2x))]dx#
Now one can use the identity #int(1/(ax+b))dx=1/aln(ax+b)#
Hence, #int[1/(3(1+x))+2/(3(1-2x))]dx#
= #1/3ln(1+x)+2/3xx(-1/2)ln(1-2x)+c#
= #1/3ln(1+x)-1/3ln(1-2x)+c#
