∫1/((x+1)^0.5+(x+1)^1.5)dx
=∫1/((x+1)^0.5+(x+1)^1.5)d(x+1)
=∫1/((x+1)^0.5(1+(x+1))d(x+1)
=∫1/((x+1)^0.5(1+(x+1))d(x+1)
=-∫1/( (1+(x+1))d(x+1)^0.5
令(x+1)^0.5=t a=1 b=3^0.5
=-∫1/( (1+t^2)dt
=-∫1/( (1+t^2)dt
=-(acrtg(t)+C)
代入a b
-acrtg(3^0.5)+acrtg(1)
=-π/3+π/4=-π/12