What is the integral of x/(2021^x) ?

The integral of x/(2021^x) is -(2021^{-x}x)/(ln(2021))-1/(ln^2(2021))*2021^{-x}+C

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integral of x/(2021^x)

\int \frac{x}{202 1 ^{x}} d x

- \frac{202 1 ^{- x} x}{ln ( 2021 )} - \frac{1}{ln ^{2} ( 2021 )} \cdot 202 1^{- x} + C

Solution steps

\int \frac{x}{202 1 ^{x}} d x

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{x}{202 1 ^{x}} = (x) 202 1^{- x}

= \int x \cdot 202 1^{- x} d x

Apply Integration By Parts

: - \frac{202 1 ^{- x} x}{ln ( 2021 )} - \int - \frac{202 1 ^{- x}}{ln ( 2021 )} d x

= - \frac{202 1 ^{- x} x}{ln ( 2021 )} - \int - \frac{202 1 ^{- x}}{ln ( 2021 )} d x

\int - \frac{202 1 ^{- x}}{ln ( 2021 )} d x = \frac{1}{ln ^{2} ( 2021 )} \cdot 202 1^{- x}

= - \frac{202 1 ^{- x} x}{ln ( 2021 )} - \frac{1}{ln ^{2} ( 2021 )} \cdot 202 1^{- x}

Add a constant to the solution

= - \frac{202 1 ^{- x} x}{ln ( 2021 )} - \frac{1}{ln ^{2} ( 2021 )} \cdot 202 1^{- x} + C

\frac{d}{d x} (x^{11} h (x))

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

f (x) = \frac{8}{x}

x \to - π lim ((x + π) cot (x))

\int sin^{7} (y) d y

What is the integral of x/(2021^x) ?
The integral of x/(2021^x) is -(2021^{-x}x)/(ln(2021))-1/(ln^2(2021))*2021^{-x}+C