不定积分（x+1)/(x^2+2x+5)dx=

Question

如题所述

Answer 1

原式=∫(x+1)dx/[(x+1)²+4]
=1/2∫d(x+1)²/[(x+1)²+4]
=1/2*ln[(x+1)²+4]+C
=1/2*ln(x²+2x+5)+C