Verified answer

Ответ:arctg³x/3 + In(x2+1)/2 +

Пошаговое объяснение:

1=f(arctg2x+x)dx/ (1+x²)=farctg²x dx/(1+x²) + Jxdx /(1+x²)

11=farctg²x dx/(1+x²) = пусть arctgx=U, = U'=dx/(1+x2) dx=(1+x2)dU

=> Тогда I = f(arctg²x+x)dx/ (1+x²)=arctg³x /3 + In(x2+1)/2 +C

11=farctg²x dx/(1+x²) =[U²dU=U³/3=arctg³x /3

12=Jxdx /(1+x2) = пусть х²+1=t, = 2x dx=dt →

dx=dt/2x =

12=fxdx /(1+x²) =(1/2) -fdt/t= (1/2) . In U = In(x2+1)/2