分部积分法,求不定积分lnx/(3x-2)^1/2

1个回答

  • 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