Question

高数两题。

求解这两题

Answer 1

先分母有理化，然后拆开，前一项直接积，后一项换元再积

追问

谢谢，另外能麻烦接一下上面写这题么

追答

求导啊，求二阶导数就行了

Answer 2

3. f'(x) = √[(x-1)^2+1] ,
f''(x) = (1/2)[(x-1)^2+1]^(-1/2) 2(x+1) = (x+1)/√[(x-1)^2+1]
f''(x) = 0 得 x = -1,
x < -1 时，f''(x) < 0; x > -1 时，f''(x) > 0,
则 x = - 1 时为拐点， f(-1) = ∫<1, -1>√[(t-1)^2+1]dt
令 t = 1+tanu， 则 dt = (secu)^2du
f(-1) = ∫<0, -arctan2>(secu)^3du = ∫<0, -arctan2>secudtanu
= [secutanu]<0, -arctan2> - ∫<0, -arctan2>(tanu)^2secudu
= -2√5 - f(1) + ∫<0, -arctan2>secudu
解得 f(-1) = -√5 + (1/2)[ln(secu+tanu)]<0, -arctan2>
= -√5 + (1/2)ln(√5-2)
得拐点 (-1, -√5+(1/2)ln(√5-2) )
曲线凸区间 (-∞, -1), 凹区间 (-1, +∞).

Answer 3