请教两个大一的高数题~前两道积分题,积分号就用 f 代替了1.f [1/(1+√1-x2）] dx2.f [ x（lnx

1.1/[1+√(1-x²)] = [1-√(1-x²)]//[1 -(1-x²)]
= [1- √(1-x²)]/x²
= 1/x² - √(1-x²)/x²
∫dx/x² - ∫√(1-x²)dx/x²
= -1/x + ∫√(1-x²)d(1/x)
= -1/x + √(1-x²)/x - ∫(1/x)d√(1-x²)
= -1/x + √(1-x²)/x - ∫(-2x)dx/[2x√(1-x²)]
= -1/x + √(1-x²)/x + ∫dx/√(1-x²)
= -1/x + √(1-x²)/x + arcsinx + C
2.
∫xln²xdx
= (1/2)∫ln²xdx²
= (1/2)(x²ln²x - ∫x²dln²x)
= (1/2)[x²ln²x - ∫x²(2lnx/x)dx]
= (1/2)(x²ln²x - ∫2xlnxdx)
= (1/2)(x²ln²x - ∫lnxdx²)
= (1/2)(x²ln²x - x²lnx + ∫x²dlnx)
= (1/2)(x²ln²x - x²lnx + ∫xdx)
= (1/2)(x²ln²x - x²lnx + x²/2)
= x²(2ln²x - 2lnx + 1)/4
3.
根据Langrange中值定理,存在t,使f'(t) = [f(1) - f(0)]/(1 - 0) = 1/e - 1
现在只需证明:在(0,1)内存在t,使-e^(-t) = 1/e - 1
显然g(t) = -e^(-t)单调
g(0) = -1
g(1) = -1/e
1/e - 1 - g(0) = 1/e - 1 + 1 = 1/e > 0,1/e - 1 > g(0)
1/e - 1 - g(1) = 1/e - 1 + 1/e = 2/e - 1 < 0,1/e - 1 < g(1)
所以在(0,1)内存在t,使-e^(-t) = 1/e - 1

哇，好复杂，我从来不想这么复杂的问题，我只想今天我的钱包有没有少钱，别人的钱包有没有掉钱