integral of (4x-1)^2

Lösung

\int (4 x - 1)^{2} d x

\frac{16 x ^{3}}{3} - 4 x^{2} + x + C

Schritte zur Lösung

\int (4 x - 1)^{2} d x

Multipliziere aus

(4 x - 1)^{2} : 16 x^{2} - 8 x + 1

= \int 16 x^{2} - 8 x + 1 d x

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 16 x^{2} d x - \int 8 x d x + \int 1 d x

\int 16 x^{2} d x = \frac{16 x ^{3}}{3}

\int 8 x d x = 4 x^{2}

\int 1 d x = x

= \frac{16 x ^{3}}{3} - 4 x^{2} + x

Füge eine Konstante zur Lösung hinzu

= \frac{16 x ^{3}}{3} - 4 x^{2} + x + C

in v erse l a pl a ce \frac{1}{s ^{2} ( s + 2 )}

\frac{d}{d x} ((lo g_{10} (x))^{2})

y^{^{''}} + 3 y = 5

y^{^{''}} + y = \frac{1}{2 k ^{2}} (1 + cos (2 t))

\frac{d}{d x} ((x^{5} + 1)^{4})