Question

高等数学求过程

Answer 1

y' = xln(1+x^2)
y=∫ xln(1+x^2) dx = 0.5∫ln(1+x^2)dx^2 = 0.5 x^2 ln(1+x^2) - 0.5 ∫x^2dln(1+x^2)
=0.5 x^2 ln(1+x^2) - 0.5 ∫x^2 *2x /(1+x^2)dx
=0.5 x^2 ln(1+x^2) - 0.5 ∫x^2 /(1+x^2)dx^2
= 0.5 x^2 ln(1+x^2) - 0.5 ∫1- 1 /毁燃(1+x^2)dx^2
= 0.5 x^2 ln(1+x^2) - 0.5 x^2 +ln(1+x^2)+C
然察哪后吧(0,-1/2)带人求纤没虚出C即可

Answer 2

y'=xln(1+x²)
y=∫xln(1+x²)dx
=1/2∫ln(1+x²)d(1+x²)
令1+x²=t
y=1/2∫lntdt
=1/2tlnt-1/2∫dt
=1/返好2tlnt-1/2t+c
=1/2(1+x²)ln(1+x²)-1/2(1+x²)+c
x=0,y=-1/2代入，者改得
-1/2=-1/2+c
c=0
所以漏嫌铅
f(x)=1/2(1+x²)ln(1+x²)-1/2(1+x²)