高数 一道题

见上图。。。

高中知识就可以解决
令f(x)=x-sinx (x≥0)
∴f'(x)=1-cosx≥0
∴f(x)在[0,+∞)上单调递增
∴f(x)min=f(0)=0
∴x-sinx≥0，即sinx≤x

(x^2+1)/[(x^2-1)(x+1)] =1/(x+1) + 2/[(x^2-1)(x+1)] let 2/[(x^2-1)(x+1)]≡ A/(x+1) +B/(x+1)^2 + C/(x-1) => 2 ≡ A(x+1)(x-1) +B(x-1) + C(x+1)^2 x=1, C=1/2 x=-1, B=-1 coef. of x^2 A+C =0 A= -1/2 2/[(x^2-1)(x+1)]≡ -(1/2)[1/(x+1)] -1/(x+1)^2 + (1/2)[1/(x-1)] (x^2+1)/[(x^2-1)(x+1)] ≡ (1/2)[1/(x+1)] -1/(x+1)^2 + (1/2)[1/(x-1)] ∫(x^2+1)/[(x^2-1)(x+1)] dx =∫ { (1/2)[1/(x+1)] -1/(x+1)^2 + (1/2)[1/(x-1)] } dx =(1/2)ln|x^2-1| +1/(x+1) + C