What is the integral of (2^x+x)^2 ?

The integral of (2^x+x)^2 is (2^{2x-1}}{ln(2)}+(2^{x+1}x)/(ln(2))-1/(ln^2(2))*2^{1+x}+\frac{x^3)/3+C

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integral of (2^x+x)^2

\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

Solution steps

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

Expand

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

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

Apply the Sum Rule:

\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}

Add a constant to the solution

= \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

\frac{d y}{d x} = 16 x sec^{2} (4 x)

\frac{d}{d x} (y sinh (x^{2}))

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

n = 0 \sum \infty n (- 1)^{n}

What is the integral of (2^x+x)^2 ?
The integral of (2^x+x)^2 is (2^{2x-1}}{ln(2)}+(2^{x+1}x)/(ln(2))-1/(ln^2(2))*2^{1+x}+\frac{x^3)/3+C