分部积分法,求不定积分lnx/(3x-2)^1/2

Question

分部积分法,求不定积分lnx/(3x-2)^1/2

Answer 1

令u=√(3x-2),得x=(u²+2)/3,dx=2u/3 du
∫(lnx)/√(3x-2) dx
=∫ln[(u²+2)/3] /u·2u/3 du
=2/3·∫ln[(u²+2)/3] du
=2/3·u·ln[(u²+2)/3]-2/3·∫2u²/(u²+2) du
=2/3·u·ln[(u²+2)/3]-4/3·∫u²/(u²+2) du
=2/3·u·ln[(u²+2)/3]-4/3·∫(u²+2-2)/(u²+2) du
=2/3·u·ln[(u²+2)/3]-4/3·∫[1-2/(u²+2)] du
=2/3·u·ln[(u²+2)/3]-4/3·∫1 du+8/3·∫1/(u²+2) du
=2/3·u·ln[(u²+2)/3]-4/3·u+2√2/3·∫1/[(u/√2)²+1] d(u/√2)
=2/3·u·ln[(u²+2)/3]-4/3·u+2√2/3·arctan(u/√2)+C