integral of (2^x+x)^2

Lösung

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

\frac{2 ^{2 x - 1}}{ln ( 2 )} + \frac{2 ^{x + 1} x}{ln ( 2 )} - \frac{1}{ln ^{2} ( 2 )} \cdot 2^{1 + x} + \frac{x ^{3}}{3} + C

Schritte zur Lösung

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

Multipliziere aus

(2^{x} + x)^{2} : 4^{x} + 2^{1 + x} x + x^{2}

= \int 4^{x} + 2^{1 + x} x + x^{2} 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 4^{x} d x + \int 2^{1 + x} x d x + \int x^{2} d x

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

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

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

= \frac{2 ^{2 x - 1}}{ln ( 2 )} + \frac{2 ^{x + 1} x}{ln ( 2 )} - \frac{1}{ln ^{2} ( 2 )} \cdot 2^{1 + x} + \frac{x ^{3}}{3}

Füge eine Konstante zur Lösung hinzu

= \frac{2 ^{2 x - 1}}{ln ( 2 )} + \frac{2 ^{x + 1} x}{ln ( 2 )} - \frac{1}{ln ^{2} ( 2 )} \cdot 2^{1 + x} + \frac{x ^{3}}{3} + C

\int cos (\frac{π x}{2}) d x

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