高数求不定积分求∫dx/[x根号(ax＋b)]

Question

高数 求不定积分
求∫dx/[x根号(ax＋b)]

Answer 1

∫dx/[x(ax+b)^(1/2)].
a=0时,必有b>0.∫dx/[x*b^(1/2)] = b^(-1/2)∫dx/x = b^(-1/2)ln|x| + c.c为任意常数.
a不为0时,令u=(ax+b)^(1/2)>0,du = adx/[2(ax+b)^(1/2)],
u^2 = ax+b,x = (u^2 - b)/a,dx = 2udu/a
∫dx/[x(ax+b)^(1/2)] = ∫2udu*a/[(u^2-b)*a*u] = 2∫du/(u^2-b).
b=0时,ax>0.
∫dx/[x(ax+b)^(1/2)]=2∫du/(u^2-b)=2∫du/u^2=-2/u + c = -2/(ax)^(1/2) + c.c 为任意常数.
b>0时,
∫dx/[x(ax+b)^(1/2)]=2∫du/(u^2-b)=b^(-1/2)∫du/[u-b^(1/2)] - b^(1/2)∫du/[u+b^(1/2)]
=b^(-1/2)[ln|u-b^(1/2)| - ln[u+b^(1/2)]] + c = b^(-1/2)[ ln|(ax+b)^(1/2)-b^(1/2)| - ln[(ax+b)^(1/2)+b^(1/2)] ] + c.c为任意常数.
b