What is the integral of 2^xx^2 ?

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

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integral of 2^xx^2

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

\int x^{5} x^{6} + 3 d x

f (x) = \frac{e ^{x}}{- 8 x ^{7}}

y^{^{'}} + y = t e^{- t} + e^{t}

x \to 0 lim (ln (\frac{x}{x - 1}))

\frac{e ^{- s}}{s - 8}

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