x/(2-x^3-3x^2)怎么对x求不定积分rt

Question

x/(2-x^3-3x^2)怎么对x求不定积分
rt

Answer 1

求不定积分∫x/(2-x³-3x²)dx
∫x/(2-x³-3x²)dx= -∫[x/(x+1)(x²+2x-2)]dx= -(1/3)∫{[1/(x+1)]-[(x-2)/(x²+2x-2)]}dx
=-(1/3)[∫dx/(x+1)-∫(x-2)dx/(x²+2x-2)]=-(1/3){ln∣x+1∣-∫[(x+2)-4]dx/(x²+2x-2)]}
=-(1/3){ln∣x+1∣-∫(x+2)dx/(x²+2x-2)+4∫dx/(x²+2x-2)}
=-(1/3){ln∣x+1∣-(1/2)∫d(x²+2x-2)/(x²+2x-2)+4∫dx/(x²+2x-2)}
=-(1/3){ln∣x+1∣-(1/2)ln∣(x²+2x-2)∣+4∫dx/[(x+1)²-3]}
=-(1/3){ln∣x+1∣-(1/2)ln∣(x²+2x-2)∣+4∫dx/[(x+1+√3)(x+1-√3)]}
=-(1/3)ln∣x+1∣+(1/6)ln∣(x²+2x-2)∣-(4/3)[1/(2√3)]∫[1/(x+1-√3)-1/(x+1+√3)]dx
=-(1/3)ln∣x+1∣+(1/6)ln∣(x²+2x-2)∣-[2/(3√3)][ln∣x+1-√3∣-ln∣x+1+√3∣]+C
=-(1/3)ln∣x+1∣+(1/6)ln∣(x²+2x-2)∣-[2/(3√3)]ln∣(x+1-√3)/(x+1+√3∣+C