Question

求微分和定积分。 上面的式子求微分，下面的求定积分

Answer 1

f(x) =(1/2)(x√(a^2-x^2) + a^2arcsin(x/a) )
f'(x) =(1/2){ -x^2/√(a^2-x^2) + √(a^2-x^2) + a^2 [1/√(1- (x/a)^2)](1/a) }
=(1/2){ -x^2/√(a^2-x^2) + √(a^2-x^2) + a^2 /√(a^2- x^2)] }
=√(a^2-x^2)

∫(0->1) (x+2)/(x^2-2x+2) dx
=(1/2)∫(0->1) (2x-2)/(x^2-2x+2) dx + 3∫(0->1) dx/(x^2-2x+2)
=(1/2)[ ln|x^2-2x+2| ]|(0->1) + 3∫(0->1) dx/(x^2-2x+2)
=-(1/2)ln2 +3∫(0->1) dx/(x^2-2x+2)

consider
x^2-2x+2 = (x-1)^2 +1
let
x-1 = tany
dx = (secy)^2 dy

∫(0->1) dx/(x^2-2x+2)
=∫(0->1) dy
=y + C'
=arctan(x-1) + C'

∫(0->1) (x+2)/(x^2-2x+2) dx
=-(1/2)ln2 +3∫(0->1) dx/(x^2-2x+2)
=-(1/2)ln2 +3arctan(x-1) + C

追问

非常感谢，第二问的arctan把上下限带进去还能化简的

Answer 2

太简单了 ＝。＝ 手上没相机没办法给你拍