求√(1+t²)的原函数即∫√(1+t²)dt=√(1+t²)*t-∫td√(1+t²)=√(1+t²) *t-∫t²/√(1+t²)dt=√(1+t²)*t-∫[(√(1+t²))²-1]/√(1+t²)dt=√(1+t²)*t-∫√...