What is the integral of-e^{-x}ln(x) ?

The integral of-e^{-x}ln(x) is e^{-x}ln(x)-\Ei(-x)+C

Popular >

integral of-e^{-x}ln(x)

\int - e^{- x} ln (x) d x

e^{- x} ln (x) - Ei (- x) + C

Solution steps

\int - e^{- x} ln (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply Integration By Parts

: - e^{- x} ln (x) - \int - \frac{e ^{- x}}{x} d x

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

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

= - (- e^{- x} ln (x) - (- Ei (- x)))

Simplify

= e^{- x} ln (x) - Ei (- x)

Add a constant to the solution

= e^{- x} ln (x) - Ei (- x) + C

(1 + x^{2}) (e^{2 t} d t - e^{x} d x) - (1 + x) d x = 0

\frac{s e ^{- 2 s}}{s ^{2} + 16}

\int_{0}^{1} cos^{2} (π x) d x

\int \frac{2 - x}{x ^{2}} d x

\frac{d}{d y} ((x^{2} + y^{2} + z^{2})^{- \frac{1}{2}})

What is the integral of-e^{-x}ln(x) ?
The integral of-e^{-x}ln(x) is e^{-x}ln(x)-\Ei(-x)+C