erweitern f(2x-1)^5dx

Lösung

erweitern

f (2 x - 1)^{5} d x

32 f x^{5} d x - 80 f x^{4} d x + 80 f x^{3} d x - 40 f x^{2} d x + 10 f x d x - fd x

Schritte zur Lösung

f (2 x - 1)^{5} d x

(2 x - 1)^{5} = 32 x^{5} - 80 x^{4} + 80 x^{3} - 40 x^{2} + 10 x - 1

= f (32 x^{5} - 80 x^{4} + 80 x^{3} - 40 x^{2} + 10 x - 1) d x

Setze Klammern

= fd x \cdot 32 x^{5} + fd x (- 80 x^{4}) + fd x \cdot 80 x^{3} + fd x (- 40 x^{2}) + fd x \cdot 10 x + fd x (- 1)

Wende Minus-Plus Regeln an

+ (- a) = - a

= 32 f x^{5} d x - 80 f x^{4} d x + 80 f x^{3} d x - 40 f x^{2} d x + 10 f x d x - 1 \cdot fd x

Multipliziere:

1 \cdot f = f

= 32 f x^{5} d x - 80 f x^{4} d x + 80 f x^{3} d x - 40 f x^{2} d x + 10 f x d x - fd x

e x p an d (1 + ln (x)) d x + (1 + l m y) d y

e x p an d \frac{40000}{( t + 2 ) ^{2}}

e x p an d \frac{1}{( s + 5 ) ( s ^{2} + 4 )}

e x p an d - e^{- 0.01 (x + h)} - 0.18 (x + h) + 7

\frac{\partial}{\partial y} ((z) (x + 2 y + 3 z))