Hallo,
Allgemein gilt:
∫ u' v dx= u v -∫ u v' dx
1.Mal:
u= - cos(x+1) v= x^2
u'= sin(x+1) v'= 2x
------------>
= - cos(x+1) *x^2 -∫ - cos(x+1) *2x dx
= - cos(x+1) *x^2 +2∫ cos(x+1) x dx
2.Mal:
∫ cos(x+1) x dx
u= sin (x+1) v= x
u'= cos(x+1) v'= 1
--->
= - cos(x+1) *x^2 +2 (sin(x+1) *x -∫ sin(x+1) *1) dx
= - cos(x+1) *x^2 +2 (sin(x+1) *x -( -cos(x+1) ))+C
= - cos(x+1) *x^2 + 2 sin(x+1) *x +2 cos(x+1) +C
Lösung mit Grenzen: π^2 -2π -2 (≈1.5864)