求y=∫(e^x,0)ln(2+t^2)dt的导数dy⼀dx

2024-12-04 22:19:06
推荐回答(3个)
回答(1):

y=∫ln(2+t²)dt=tln(2+t²)-∫td(ln(2+t²))=tln(2+t²)-∫t*2t/(2+t²)dt=tln(2+t²)-∫2(t²+2)/(2+t²)-4/(2+t²)dt=tln(2+t²)-2t+∫(2√2)/(1+(t/√2)²)d(t/√2)=tln(2+t²)-2t+2√2arctan(t/√2)把(e∧x,0)代入得y=(e∧x)ln(2+e∧(2x))-2e∧x+2√2arctan((e∧x)/√2)
则dy/dx=

回答(2):

回答(3):


如图