Question

这几道高数怎么做？

Answer 1

令t=x^2＞0(x＞0），则x=√t，于是f'（t）=1+√t，从而f'（x）=1+√x，两端积分得
∫f'（x）dx=∫(1+√x)dx，f(x)=x+(2/3)x√x+C，

Answer 2

1) f'(x) = 1+sqrt(x)
f(x) = x + (2/3)x^(3/2) + c
2) f'(x) = sqrt(1-x^2) + x/(1-x)
f(x) = (1/2)[arcsin(x) + xsqrt(1-x^2)] - x - ln|1-x| + c