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

The integral from 1 to e^2 of xln(x) is (3e^4+1)/4

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

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

\frac{3 e ^{4} + 1}{4}

Decimal Notation

41.19861 \dots

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{x}{2} d x = \frac{x ^{2}}{4}

= [\frac{1}{2} (x^{2} ln (x) - 2 \cdot \frac{x ^{2}}{4})]_{1}^{e^{2}}

2 \cdot \frac{x ^{2}}{4} = \frac{x ^{2}}{2}

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

Compute the boundaries:

\frac{3 e ^{4}}{4} + \frac{1}{4}

= \frac{3 e ^{4}}{4} + \frac{1}{4}

Simplify

= \frac{3 e ^{4} + 1}{4}

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What is the integral from 1 to e^2 of xln(x) ?
The integral from 1 to e^2 of xln(x) is (3e^4+1)/4