=x*arctgx - ∫ xdx/(1+x^2)=
=x*arctgx -(1/2) ln(1+x^2)+C_(1)
y= ∫(x*arctgx -(1/2) ln(1+x^2)+C_(1))=
= [b] ∫ x*arctgxdx - (1/2) ∫ ln(1+x^2)dx+ C_(1) ∫ dx[/b]
Считаем каждый интеграл отдельно:
∫ x*arctgxdx [по частям: u=arctgx; dv=xdx ⇒ u=dx/(1+x^2); v=x^2/2]=
=(1/2)x^2arctgx-(1/2) ∫x^2dx/(1+x^2)=
=(1/2)x^2arctgx-(1/2) ∫(x^2+1-1)dx/(1+x^2)=
=(1/2)x^2arctgx-(1/2) ∫dx +(1/2)∫dx/(1+x^2)=
= (1/2)x^2arctgx-(1/2)*x +(1/2)arctgx
∫ ln(1+x^2)dx= [по частям: u=ln(1+x^2); dv=dx ⇒du=2xdx/(1+x^2); v=x]
=x*ln(1+x^2)- ∫ x*(2dx)/(1+x^2)=
=x*ln(1+x^2) -2 ∫x^2dx/(1+x^2)=
=x*ln(1+x^2)-2 ∫(x^2+1-1)dx/(1+x^2)=
=x*ln(1+x^2)-2 ∫ dx+2 ∫ dx/(1+x^2)
=x*ln(1+x^2)-2x+2arctgx
Итак
∫ x*arctgxdx - (1/2) ∫ ln(1+x^2)dx+ C=
=(1/2)x^2arctgx-(1/2)*x +(1/2)arctgx-(1/2)x*ln(1+x^2)+x-arctgx +C_(1)x+C_(2)
y=(1/2)x^2arctgx+(1/2)*x -(1/2)arctgx-(1/2)x*ln(1+x^2)+C_(1)x+C_(2)- общее решение
y`=x*arctgx -(1/2) ln(1+x^2)+C_(1)
Из условий:
y(0)=0
y`(0)=0
0=С_(2)
0=С_(1)
y=(1/2)x^2arctgx+(1/2)*x -(1/2)arctgx-(1/2)x*ln(1+x^2) - частное решение