What is the integral from 1 to infinity of x/(2^x) ?

The integral from 1 to infinity of x/(2^x) is -(-1/(2ln(2))-1/(2ln^2(2)))

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integral from 1 to infinity of x/(2^x)

\int_{1}^{\infty} \frac{x}{2 ^{x}} d x

- (- \frac{1}{2 ln ( 2 )} - \frac{1}{2 ln ^{2} ( 2 )})

Decimal Notation

1.76203 \dots

Solution steps

\int_{1}^{\infty} \frac{x}{2 ^{x}} d x

Compute the indefinite integral:

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

Compute the boundaries:

\int_{1}^{\infty} \frac{x}{2 ^{x}} d x = 0 - (- \frac{1}{2 ln ( 2 )} - \frac{1}{2 ln ^{2} ( 2 )})

= 0 - (- \frac{1}{2 ln ( 2 )} - \frac{1}{2 ln ^{2} ( 2 )})

Simplify

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

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What is the integral from 1 to infinity of x/(2^x) ?
The integral from 1 to infinity of x/(2^x) is -(-1/(2ln(2))-1/(2ln^2(2)))