What is the integral of x^2*2^{-x} ?

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

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integral of x^2*2^{-x}

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

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

Solution steps

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

Apply u-substitution

: \int - 2^{u} u^{2} d u

= \int - 2^{u} u^{2} d u

Take the constant out:

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

= - \int 2^{u} u^{2} d u

Apply Integration By Parts

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

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

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

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

Substitute back

u = - 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}))

Simplify

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

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

Add a constant to the solution

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

\int x e^{- 2} d x

f (x) = 4 + 8 x - 5 x^{2}

e^{e^{x^{2}}}

(x^{3} - y^{3}) d x + (x y^{2}) d y = 0

\frac{d y}{d x} = 3 (y + 7) x^{2}

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