integral of 4x(4x+1)^3

Lösung

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

4 (\frac{64 x ^{5}}{5} + 12 x^{4} + 4 x^{3} + \frac{x ^{2}}{2}) + C

Schritte zur Lösung

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

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int x (4 x + 1)^{3} d x

Multipliziere aus

x (4 x + 1)^{3} : 64 x^{4} + 48 x^{3} + 12 x^{2} + x

= 4 \cdot \int 64 x^{4} + 48 x^{3} + 12 x^{2} + x 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

= 4 (\int 64 x^{4} d x + \int 48 x^{3} d x + \int 12 x^{2} d x + \int x d x)

\int 64 x^{4} d x = \frac{64 x ^{5}}{5}

\int 48 x^{3} d x = 12 x^{4}

\int 12 x^{2} d x = 4 x^{3}

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

= 4 (\frac{64 x ^{5}}{5} + 12 x^{4} + 4 x^{3} + \frac{x ^{2}}{2})

Füge eine Konstante zur Lösung hinzu

= 4 (\frac{64 x ^{5}}{5} + 12 x^{4} + 4 x^{3} + \frac{x ^{2}}{2}) + C

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

\int 3 x^{2} + e^{- x} d x

\int e^{3 x} sin (e^{3 x}) d x

\int \frac{x + 1}{x x - 2} d x

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