What is the integral from 1 to 2 of e^xln(x) ?

The integral from 1 to 2 of e^xln(x) is e^2ln(2)-\Ei(2)+\Ei(1)

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integral from 1 to 2 of e^xln(x)

\int_{1}^{2} e^{x} ln (x) d x

e^{2} ln (2) - Ei (2) + Ei (1)

Solution steps

\int_{1}^{2} e^{x} ln (x) d x

Apply Integration By Parts

: [e^{x} ln (x) - \int \frac{e ^{x}}{x} d x]_{1}^{2}

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

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

= [e^{x} ln (x) - Ei (x)]_{1}^{2}

Compute the boundaries:

e^{2} ln (2) - Ei (2) + Ei (1)

= e^{2} ln (2) - Ei (2) + Ei (1)

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What is the integral from 1 to 2 of e^xln(x) ?
The integral from 1 to 2 of e^xln(x) is e^2ln(2)-\Ei(2)+\Ei(1)