integral of 36(4x+4)^2-2sqrt(3x-1)

Lösung

\int 36 (4 x + 4)^{2} - 2 3 x - 1 d x

36 (16 x^{3} - \frac{32 x ^{3}}{3} + 16 x^{2} + 16 x) - \frac{4}{9} (3 x - 1)^{\frac{3}{2}} + C

Schritte zur Lösung

\int 36 (4 x + 4)^{2} - 2 3 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 36 (4 x + 4)^{2} d x - \int 2 3 x - 1 d x

\int 36 (4 x + 4)^{2} d x = 36 (16 x^{3} - \frac{32 x ^{3}}{3} + 16 x^{2} + 16 x)

\int 2 3 x - 1 d x = \frac{4}{9} (3 x - 1)^{\frac{3}{2}}

= 36 (16 x^{3} - \frac{32 x ^{3}}{3} + 16 x^{2} + 16 x) - \frac{4}{9} (3 x - 1)^{\frac{3}{2}}

Füge eine Konstante zur Lösung hinzu

= 36 (16 x^{3} - \frac{32 x ^{3}}{3} + 16 x^{2} + 16 x) - \frac{4}{9} (3 x - 1)^{\frac{3}{2}} + C

\int 2 csc (2 x) d x

x \to 0 + lim (\frac{∣ x - 4 ∣}{x - 4})

\int (3 x^{2} - 8 x + 3) d x

\int x^{2} 7 x - 5 d x

t^{3} \frac{d y}{d t} + 4 t^{2} y = e^{- t}