∫[1/(2+1√(x+1))]dx＝？
∫e^x(1/x)dx=∫(1/x)de^x=e^x(1/x)-∫e^xd(1/x)=e^x(1/x)+∫e^x(1/x^2)dx 故∫e^x(1/x-1/x^2)dx=e^x(1/x)+c
![∫[1/(2+1√(x+1))]dx＝？](https://image.zuidongwo.com/picture/pic8306.jpg)
∫x/√(1-x^2)dx=1/2∫1/√(1-x^2)dx^2=-1/2∫1/√(1-x^2)d(1-x^2)=-√(1-x^2)+c
如图所示:
∫e^x(1/x)dx=∫(1/x)de^x=e^x(1/x)-∫e^xd(1/x)=e^x(1/x)+∫e^x(1/x^2)dx 故∫e^x(1/x-1/x^2)dx=e^x(1/x)+c
∫x/√(1-x^2)dx=1/2∫1/√(1-x^2)dx^2=-1/2∫1/√(1-x^2)d(1-x^2)=-√(1-x^2)+c
如图所示: