∫ln(x+1)/√(x+1)=dx
令x+1=t2,则x=t2-1,dx=2tdt, ∫[ln(x+1)/√(x+1)]dx=4∫lntdt =4tlnt-4∫tdlnt=4tlnt-4t+C =4√(x+1)ln √(x+1)-4√(x+1)+C =2√(x+1)ln(x+1)-4√(x+1)+C
∫ln(x+1)/√(x+1)=dx
令x+1=t2,则x=t2-1,dx=2tdt, ∫[ln(x+1)/√(x+1)]dx=4∫lntdt =4tlnt-4∫tdlnt=4tlnt-4t+C =4√(x+1)ln √(x+1)-4√(x+1)+C =2√(x+1)ln(x+1)-4√(x+1)+C