What is the integral of (ln(x)+1/x)e^x ?

The integral of (ln(x)+1/x)e^x is e^xln(x)+C

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integral of (ln(x)+1/x)e^x

\int (ln (x) + \frac{1}{x}) e^{x} d x

e^{x} ln (x) + C

Solution steps

\int (ln (x) + \frac{1}{x}) e^{x} d x

Multiply by the conjugate of

ln (x) + \frac{1}{x} : \frac{ln ^{2} ( x ) - \frac{1}{x ^{2}}}{ln ( x ) - \frac{1}{x}}

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

Simplify

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

= \int \frac{e ^{x} ( x ln ( x ) + 1 )}{x} d x

Expand

\frac{e ^{x} ( x ln ( x ) + 1 )}{x} : e^{x} ln (x) + \frac{e ^{x}}{x}

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

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

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

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

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

Simplify

= e^{x} ln (x)

Add a constant to the solution

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

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What is the integral of (ln(x)+1/x)e^x ?
The integral of (ln(x)+1/x)e^x is e^xln(x)+C