What is the integral of (x-1)4^x ?

The integral of (x-1)4^x is (2^{2x-1}x)/(ln(2))-1/(ln^2(2))*2^{2x-2}-(2^{2x-1})/(ln(2))+C

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integral of (x-1)4^x

\int (x - 1) 4^{x} d x

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

Solution steps

\int (x - 1) \cdot 4^{x} d x

Expand

(x - 1) \cdot 4^{x} : 4^{x} x - 4^{x}

= \int 4^{x} x - 4^{x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 4^{x} x d x - \int 4^{x} d x

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

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

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

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