Question

x/(ax+b)dx的不定积分正推导怎么推?x^2/(ax+b)dx呢?

Answer 1

x/(ax+b)dx=1/a^2*ax/(ax+b)dax=1/a^2(1-b/(ax+b))dax=1/a^2*d(ax+b)-b/a^2*d(ax+b)/(ax+b)

积分得(ax+b)/a^2-b/a^2*ln(ax+b)

x^2/(ax+b)dx化成1/a^3*((ax+b)^2-b^2-2abx)/(ax+b)dax ,以下计算跟第一中情况基本一样