What is the integral of x4^x ?

The integral of x4^x is (2^{2x-1}x)/(ln(2))-1/(ln^2(2))*2^{2x-2}+C

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integral of x4^x

\int x 4^{x} d x

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

Solution steps

\int x \cdot 4^{x} d x

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

y = (cos (x))^{s i n (x)}

\int \frac{x}{2 x + 1} d x

x \to 2 lim ((3 - x) (x + 1))

\frac{\partial}{\partial x} (\frac{5}{x y} - 4 x^{2} - 2 y^{2})

\int \frac{1}{x ^{2} - 4 x + 8} d x

What is the integral of x4^x ?
The integral of x4^x is (2^{2x-1}x)/(ln(2))-1/(ln^2(2))*2^{2x-2}+C