What is the integral of x^32^x ?

The integral of x^32^x is 1/(ln^4(2))(ln^3(2)*2^xx^3-3(ln^2(2)*2^xx^2-2(ln(2)*2^xx-2^x)))+C

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integral of x^32^x

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

\frac{1}{ln ^{4} ( 2 )} (ln^{3} (2) \cdot 2^{x} x^{3} - 3 (ln^{2} (2) \cdot 2^{x} x^{2} - 2 (ln (2) \cdot 2^{x} x - 2^{x}))) + C

Solution steps

\int x^{3} \cdot 2^{x} d x

Apply u-substitution

: \int \frac{ln ^{3} ( u )}{ln ^{4} ( 2 )} d u

= \int \frac{ln ^{3} ( u )}{ln ^{4} ( 2 )} d u

Take the constant out:

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

= \frac{1}{ln ^{4} ( 2 )} \cdot \int ln^{3} (u) d u

Apply Integration By Parts

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

= \frac{1}{ln ^{4} ( 2 )} (u ln^{3} (u) - \int 3 ln^{2} (u) d u)

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

= \frac{1}{ln ^{4} ( 2 )} (u ln^{3} (u) - 3 (u ln^{2} (u) - 2 (u ln (u) - u)))

Substitute back

u = 2^{x}

= \frac{1}{ln ^{4} ( 2 )} (2^{x} ln^{3} (2^{x}) - 3 (2^{x} ln^{2} (2^{x}) - 2 (2^{x} ln (2^{x}) - 2^{x})))

Simplify

\frac{1}{ln ^{4} ( 2 )} (2^{x} ln^{3} (2^{x}) - 3 (2^{x} ln^{2} (2^{x}) - 2 (2^{x} ln (2^{x}) - 2^{x}))) : \frac{1}{ln ^{4} ( 2 )} (ln^{3} (2) \cdot 2^{x} x^{3} - 3 (ln^{2} (2) \cdot 2^{x} x^{2} - 2 (ln (2) \cdot 2^{x} x - 2^{x})))

= \frac{1}{ln ^{4} ( 2 )} (ln^{3} (2) \cdot 2^{x} x^{3} - 3 (ln^{2} (2) \cdot 2^{x} x^{2} - 2 (ln (2) \cdot 2^{x} x - 2^{x})))

Add a constant to the solution

= \frac{1}{ln ^{4} ( 2 )} (ln^{3} (2) \cdot 2^{x} x^{3} - 3 (ln^{2} (2) \cdot 2^{x} x^{2} - 2 (ln (2) \cdot 2^{x} x - 2^{x}))) + C

x \to 0 lim (sin (\frac{6}{x}))

\frac{\partial}{\partial u} (u^{2})

\frac{\partial}{\partial z} (e^{y} + x cos (x z))

f (x) = (x - 5 x x)

y^{^{'}} = 2 x + y, y (0) = 6

What is the integral of x^32^x ?
The integral of x^32^x is 1/(ln^4(2))(ln^3(2)*2^xx^3-3(ln^2(2)*2^xx^2-2(ln(2)*2^xx-2^x)))+C