不定积分小问题求2^x*3^x/(3^2x-2^2x)的原函数,用不定积分来求,麻烦写出过程,

设t=(3/2)^x,则dt=(3/2)^xln（3/2)dt∴原式=∫2^x*3^x/(3^2x-2^2x)dx =1/ln(3/2)∫dt/(t²-1) =1/[2ln(3/2)]∫[1/(t-1)-1/(t+1)]dt =1/[2ln(3/2)](ln|t-1|-ln|t+1|)+C =1/[2ln(3/2)]ln|(t-1...