Question

高数，谁会？红笔写的

Answer 1

原式=∫1/(x²+2x+5) d(x²+2x+5)+∫(x+2)/(x²+2x+5) dx
=ln(x²+2x+5)+1/2 ∫[(2x+2)+2]/(x²+2x+5) dx
=ln(x²+2x+5)+1/2 ∫1/(x²+2x+5) d(x²+2x+5)+ ∫1/(x²+2x+5)dx
=ln(x²+2x+5)+1/2 ln(x²+2x+5)+ ∫1/[(x+1)²+4]dx
=3/2ln(x²+2x+5)+ 1/2 arctan(x+1)/2+C

Answer 2

∫[(3x+4)/(x²+2x+5)]dx
=(3/2)∫[(2x+6)/(x²+2x+5)]dx
=(3/2)∫[(2x+2+4)/(x²+2x+5)]dx
=(3/2)∫[(2x+2)/(x²+2x+5)]dx+(3/2)∫[4/(x²+2x+5)]dx
=(3/2)∫[1/(x²+2x+5)]d(x²+2x+5)+6∫[1/(x²+2x+5)]dx
=(3/2)ln(x²+2x+5)+6∫[1/(x+1)²+2²]dx
=(3/2)ln(x²+2x+5)+6×(1/2)arctan[(x+2)/2]+C
=(3/2)ln(x²+2x+5)+3arctan[(x+2)/2]+C