求∫1/(X+1）（x+3） dx 不定积分求解的详细过程

∫1/(X+1）（x+3） dx
=∫dx/2(x+1)-∫dx/2(x+3)
=1/2*∫d(x+1)/(x+1)-1/2*∫d(x+3)/(x+3)
=ln(x+1)/2-ln(x+3)/2+C
化简自已做追问

怎样化简

追答

ln(x+1)/2-ln(x+3)/2+C
=1/2*[ln(x+1)-ln(x+3)]+C
=ln[(x+1)/(x+3)]/2+C 注意分母2不在ln内，在ln外

本回答被提问者采纳

1/[x(x+k)]=(1/k)[1/n-1/(n+k)]
1/[(x+1)(x+3)]=(1/2)[1/(x+1)-1/(x+3)]
∫1/(X+1)(x+3) dx
=(1/2)∫[1/(x+1)-1/(x+3)] dx
=(1/2)[ln|x+1|-ln|x+3|+C]
=ln√[C(x+1)/(x+3)]