What is the integral of (e^{-x}(3+4xln(x)))/(2x) ?

The integral of (e^{-x}(3+4xln(x)))/(2x) is 1/2 (7\Ei(-x)-4e^{-x}ln(x))+C

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integral of (e^{-x}(3+4xln(x)))/(2x)

\int \frac{e ^{- x} ( 3 + 4 x ln ( x ) )}{2 x} d x

\frac{1}{2} (7 Ei (- x) - 4 e^{- x} ln (x)) + C

Solution steps

\int \frac{e ^{- x} ( 3 + 4 x ln ( x ) )}{2 x} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{e ^{- x} ( 3 + 4 x ln ( x ) )}{x} d x

Expand

\frac{e ^{- x} ( 3 + 4 x ln ( x ) )}{x} : \frac{3 e ^{- x}}{x} + 4 e^{- x} ln (x)

= \frac{1}{2} \cdot \int \frac{3 e ^{- x}}{x} + 4 e^{- x} ln (x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int \frac{3 e ^{- x}}{x} d x + \int 4 e^{- x} ln (x) d x)

\int \frac{3 e ^{- x}}{x} d x = 3 Ei (- x)

\int 4 e^{- x} ln (x) d x = 4 (- e^{- x} ln (x) + Ei (- x))

= \frac{1}{2} (3 Ei (- x) + 4 (- e^{- x} ln (x) + Ei (- x)))

Simplify

\frac{1}{2} (3 Ei (- x) + 4 (- e^{- x} ln (x) + Ei (- x))) : \frac{1}{2} (7 Ei (- x) - 4 e^{- x} ln (x))

= \frac{1}{2} (7 Ei (- x) - 4 e^{- x} ln (x))

Add a constant to the solution

= \frac{1}{2} (7 Ei (- x) - 4 e^{- x} ln (x)) + C

\int a - b x d x

\int \frac{sin ( n t )}{n} d t

\frac{\partial}{\partial x} (3 x^{2} + 2 x y^{3})

\int ln (x - 9) d x

f (x) = 3 x^{4}, at x = - 2

What is the integral of (e^{-x}(3+4xln(x)))/(2x) ?
The integral of (e^{-x}(3+4xln(x)))/(2x) is 1/2 (7\Ei(-x)-4e^{-x}ln(x))+C