∫x^5/√(1+x²)dx.令x=tany，dx=sec²ydy=∫tan^5ysecydy=∫(sec⁴y-2sec²y+1)d(secy)=(1/5)sec^5y-(2/3)sec³y+secy+C=(1/5)(1+x²)^(5/2)-(2/3)(1+x²)^(3/2)+√(1+x²)+C=(1/15)(3x⁴-4x²+8)√(1+x²)+C
