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

The integral from 1 to x^2 of e^tln(t) is 2e^{x^2}ln(x)-\Ei(x^2)+\Ei(1)

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integral from 1 to x^2 of e^tln(t)

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

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

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

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