Question

有理式不定积分∫x∧4／（1＋x∧2）∧2dx

Answer 1

∫x^4/(1+x²)² dx
=∫[1+1/(1+x²)²-2/(1+x²)]dx,用综合除法
=∫dx+∫dx/(1+x²)²-2∫dx/(1+x²)
在第二项,令x=tanp,dx=sec²pdp
=∫dx+∫sec²p/(1+tan²p)²-2∫dx/(1+x²)
=∫dx+∫sec²p/(sec^4p)-2∫dx/(1+x²)
=∫dx+∫cos²pdp-2∫dx/(1+x²)
=∫dx+∫(1+cos2p)/2 dp-2∫dx/(1+x²)
=∫dx+(1/2)∫dp+(1/4)∫cos2pd(2p)-2∫dx/(1+x²)
=x+(1/2)p+(1/4)sin2p-2arctanx+C
=x+(1/2)p+(1/2)sinpcosp-2arctanx+C
=x+(1/2)arctanx+(1/2)[x/√(1+x²)][1/√(1+x²)]-2arctanx+C
=x-(3/2)arctanx+(1/2)[x/(1+x²)]
=x+x/[2(1+x²)]-(3/2)arctanx+C