Question

f(x)=∫ln(2+t)dt,上限x^2,下线0.求f'(x).08年数学一的第一道题

Answer 1

∫ln(2+t)dt,d(2+t)=dt

=∫ln(2+t)d(2+t),分部积分法

=(2+t)ln(2+t)-∫(2+t)d[ln(2+t)]

=(2+x²)ln(2+x²)-2ln2-∫dt

=(2+x²)ln(2+x²)-2ln2-t

=(2+x²)ln(2+x²)-2ln2-x²

∴f'(x)

=d/dx (2+x²)ln(2+x²)-2ln2-x²

=ln(2+x²)*(2x)+(2+x²)*2x/(2+x²)-2x

=2xln(2+x²)+2x-2x

=2xln(2+x²)