Question

f(x)+连续函数,且+∫_0^xf(t)dt=x^3+ln(x+1)+则+f(x)=

Answer 1

问：f(x)+连续函数,且+∫_0^xf(t)dt=x^3+ln(x+1)+则+f(x)=

答：你好，很高兴为您服或册务f(x)+连续函数,且+∫_0^xf(t)dt=x^3+ln(x+1)+则+f(x)=两边求导xf(x)=2x+f'(x)设f(x)=yxy=2x+y'y'=x(y-2)dy/(y-2)=xdx两边握团滚积分lin（y-2)=1/2*x^2y-2=e^(1/2*x^2)y=e^(1/2*x^2)+2即段余 f(x)=e^(1/2*x^2)+2

答：设∫\x0910f(t)dt＝a，则f（x）=x+2a∴∫哪桐悉\x0910f(x)=∫\x0910(x+2a)dx＝1/2+2a而∫\x0910f(t)dt=∫轮伍\x0910f(x)dx＝李乎a∴1/2+2a＝a∴a＝−1/2∴y=x-1