What is the integral from 1 to 2 of e^{-x}*ln(x) ?

The integral from 1 to 2 of e^{-x}*ln(x) is -1/(e^2)ln(2)+\Ei(-2)-\Ei(-1)

Popular >

integral from 1 to 2 of e^{-x}*ln(x)

\int_{1}^{2} e^{- x} \cdot ln (x) d x

- \frac{1}{e ^{2}} ln (2) + Ei (- 2) - Ei (- 1)

Solution steps

\int_{1}^{2} e^{- x} ln (x) d x

Apply Integration By Parts

: [- e^{- x} ln (x) - \int - \frac{e ^{- x}}{x} d x]_{1}^{2}

= [- e^{- x} ln (x) - \int - \frac{e ^{- x}}{x} d x]_{1}^{2}

\int - \frac{e ^{- x}}{x} d x = - Ei (- x)

= [- e^{- x} ln (x) - (- Ei (- x))]_{1}^{2}

Simplify

= [- e^{- x} ln (x) + Ei (- x)]_{1}^{2}

Compute the boundaries:

- \frac{1}{e ^{2}} ln (2) + Ei (- 2) - Ei (- 1)

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

\int_{0}^{\frac{π}{2}} x cos (5 x) d x

\int_{0}^{4} \frac{4 - x}{( x + 2 ) ^{2}} d x

\int_{0}^{2} 44 z^{2} ln (z) d z

\int_{0}^{3} x (3 - x) d x

\int_{- 2}^{3} (x^{2} + x) d x

What is the integral from 1 to 2 of e^{-x}*ln(x) ?
The integral from 1 to 2 of e^{-x}*ln(x) is -1/(e^2)ln(2)+\Ei(-2)-\Ei(-1)