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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

f (x) = 4 x e^{x}, at x = 0

\int \frac{x ^{5} - x ^{4} + 4 x ^{3} - 4 x ^{2} + 8 x - 4}{( x ^{2} + 2 ) ^{3}} d x

- \frac{6 x}{( x ^{2} - 4 ) ^{2}}

\int ln (x - 1 + x^{2}) d x

\frac{\partial}{\partial x} (x arctan (y))

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