询求一道不定积分题详解

∫(dx/(4+9x^2 ))
=（1/4）∫(dx/(1+9/4x^2 ))
=（1/4）∫(dx/(1+u^2 ))*2/3(u=3/2x)
=1/6arctanu+C=1/6arctan3/2x+C

9x^2+c

∫(dx/(4+9x^2 ))
=(1/6)∫d(3x/2)/((3x/2)²+1)
=(1/6)arctan(3x/2)+C

(arctanx)'=1/(x²+1)
∫dx/(x²+1)=arctanx+C

通过换元，令t=3x/2即可

令x=(2/3)tana
9x^2=4(tana)^2
dx=(2/3)(seca)^2da
a=arctan(3x/2)

原式=∫(2/3)(seca)^2/[4+4(tana)^2]da
=∫(2/3)(seca)^2/4(seca)^2da
=(1/6)∫da
=a/6+C
=[arctan(3x/2)]/6+C

化成∫(dx/(a^2+x^2 ))这种格式,一查公式表就得结果了.