Question

各位大大帮忙解一下不定积分的题目！！

Answer 1

给你提供个思路
x^3/(4+x^2)dx=[(x^2+4)-4]/[2(x^2+4)] *d(x^2+4),
令x^2+4=t,即 1/2dt-(2/t) *dt =1/2t-2lnt+C,
即x^2/2-2ln(x^2+4)+C

Answer 2

原式=1/2∫x²dx²/(4+x²)
=1/2∫[1-4/(4+x²)]dx²
=1/2∫[1-4/(4+x²)]d(4+x²)
=1/2*(4+x²)-2ln|4+x²|+C