Question

设函数y=1-x/1+x e√x,求y′x=4

1、求由方程y=1+x^2-xe^y所确定的隐函数y的导数dy/dx.
2、求∫sin^2(x/2)dx.
3、计算∫1/√(2gx)dx,其中g为一正常数.
4、计算∫(√x-1)(x+1/√x)dx.

Answer 1

1、y=1+x^2-xe^y
==>y′=2x-e^y-xe^yy′
==>(1+xe^y)y′=(2x-e^y)
==>y′=(2x-e^y)/(1+xe^y)
2、∫sin^2(x/2)dx
=∫(1-cosx)/2dx
=(x+sinx)2+C (C是积分常数)
3、∫1/√(2gx)dx
=(1/√(2g))∫dx/√x
=√(2/g)∫d(√x)
=√(2x/g)+C (C是积分常数)
4、∫(√x-1)(x+1/√x)dx
=2∫(t-1)(t²+1/t)tdt (令√x=t)
=2∫(t-1)(t³+1)dt
=2∫(t^4-t³+t-1)dt
=2(t^5/5-t^4/4+t²/2-t)+C
=2[x^(5/2)/5-x²/4+x/2-√x]+C (C是积分常数)