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

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

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

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

e^{x} ln (x) + C

Solution steps

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

Multiply by the conjugate of

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

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

Simplify

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

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

Expand

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

= \int \frac{e ^{x}}{x} + 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

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

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

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

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

Simplify

= e^{x} ln (x)

Add a constant to the solution

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

\int \frac{1}{( x ^{2} + 4 x + 7 ) ^{3}} d x

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

\frac{1}{1 + x ^{2}}

\int \frac{x}{36 x ^{4} + 9} d x

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