Question

用换元法求 ∫dx/(4-9x^2)

Answer 1

令t=3x,则

原式=dx/((2+3x)(2-3x))=(dx/(1/(2+3x)+1/(2-3x)))/4=(dx/(2+3x)+dx/(2-3x))/4

=(d(3x)/(2+3x)+d(3x)/(2-3x))/(4*3)=(dt/(2+t)+dt/(2-t))/12

=(ln(t+2)-ln(2-t))/12

=(ln(3x+2)-ln(2-3x))/12