Question

求解∫(1-4t)/(12t^2+3)=?

Answer 1

求不定积分∫[(1-4t)/(12t²+3)]dt=?

原式=∫dt/(12t²+3)-4∫tdt/(12t²+3)=(1/3)∫dt/(4t²+1)-(4/3)∫tdt/(4t²+1)

=(1/6)∫d(2t)/[1+(2t)²]-(1/6)∫d(4t²+1)/(4t²+1)=(1/6)arctan(2t)-(1/6)ln(4t²+1)+C