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

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

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integral of (x^2)/(2^x)

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

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

Solution steps

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

Apply u-substitution

: \int \frac{ln ^{2} ( u )}{ln ^{3} ( 2 ) u ^{2}} d u

= \int \frac{ln ^{2} ( u )}{ln ^{3} ( 2 ) u ^{2}} d u

Take the constant out:

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

= \frac{1}{ln ^{3} ( 2 )} \cdot \int \frac{ln ^{2} ( u )}{u ^{2}} d u

Apply Integration By Parts

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

= \frac{1}{ln ^{3} ( 2 )} (- \frac{ln ^{2} ( u )}{u} - \int - \frac{2 ln ( u )}{u ^{2}} d u)

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

= \frac{1}{ln ^{3} ( 2 )} (- \frac{ln ^{2} ( u )}{u} - (- 2 (- \frac{ln ( u )}{u} - \frac{1}{u})))

Substitute back

u = 2^{x}

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} ((2 + 4 y^{2}) x - 3 x^{2})

\int \frac{x ^{2} + 5}{x} d x

\frac{x ^{2}}{5 + x}

y = \frac{14}{1 + e ^{- x}}, (0, 7)

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