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

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

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

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

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

Decimal Notation

4.19452 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

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

= \frac{e ^{2}}{2} + \frac{1}{2}

Simplify

= \frac{e ^{2} + 1}{2}

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