What is the integral of (6-x)6^x ?

The integral of (6-x)6^x is (6^{x+1})/(ln(6))-(6^xx)/(ln(6))+(6^x)/(ln^2(6))+C

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integral of (6-x)6^x

\int (6 - x) 6^{x} d x

\frac{6 ^{x + 1}}{ln ( 6 )} - \frac{6 ^{x} x}{ln ( 6 )} + \frac{6 ^{x}}{ln ^{2} ( 6 )} + C

Solution steps

\int (6 - x) \cdot 6^{x} d x

Expand

(6 - x) \cdot 6^{x} : 6^{1 + x} - 6^{x} x

= \int 6^{1 + x} - 6^{x} 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 6^{1 + x} d x - \int 6^{x} x d x

\int 6^{1 + x} d x = \frac{6 ^{x + 1}}{ln ( 6 )}

\int 6^{x} x d x = \frac{6 ^{x} x}{ln ( 6 )} - \frac{6 ^{x}}{ln ^{2} ( 6 )}

= \frac{6 ^{x + 1}}{ln ( 6 )} - (\frac{6 ^{x} x}{ln ( 6 )} - \frac{6 ^{x}}{ln ^{2} ( 6 )})

Simplify

= \frac{6 ^{x + 1}}{ln ( 6 )} - \frac{6 ^{x} x}{ln ( 6 )} + \frac{6 ^{x}}{ln ^{2} ( 6 )}

Add a constant to the solution

= \frac{6 ^{x + 1}}{ln ( 6 )} - \frac{6 ^{x} x}{ln ( 6 )} + \frac{6 ^{x}}{ln ^{2} ( 6 )} + C

What is the integral of (6-x)6^x ?
The integral of (6-x)6^x is (6^{x+1})/(ln(6))-(6^xx)/(ln(6))+(6^x)/(ln^2(6))+C