What is the integral from 1 to e of (x^2-2x+3)ln(x) ?

The integral from 1 to e of (x^2-2x+3)ln(x) is (4e^3-9e^2+47)/(18)

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integral from 1 to e of (x^2-2x+3)ln(x)

\int_{1}^{e} (x^{2} - 2 x + 3) ln (x) d x

\frac{4 e ^{3} - 9 e ^{2} + 47}{18}

Decimal

3.38003 \dots

Solution steps

\int_{1}^{e} (x^{2} - 2 x + 3) ln (x) d x

Apply Integration By Parts

: [ln (x) (\frac{x ^{3}}{3} - x^{2} + 3 x) - \int \frac{1}{3} (x^{2} - 3 x + 9) d x]_{1}^{e}

= [ln (x) (\frac{x ^{3}}{3} - x^{2} + 3 x) - \int \frac{1}{3} (x^{2} - 3 x + 9) d x]_{1}^{e}

\int \frac{1}{3} (x^{2} - 3 x + 9) d x = \frac{x ^{3}}{9} - \frac{x ^{2}}{2} + 3 x

= [ln (x) (\frac{x ^{3}}{3} - x^{2} + 3 x) - (\frac{x ^{3}}{9} - \frac{x ^{2}}{2} + 3 x)]_{1}^{e}

Simplify

[ln (x) (\frac{x ^{3}}{3} - x^{2} + 3 x) - (\frac{x ^{3}}{9} - \frac{x ^{2}}{2} + 3 x)]_{1}^{e} : [\frac{1}{18} (6 x^{3} ln (x) - 18 x^{2} ln (x) + 54 x ln (x) - 2 x^{3} + 9 x^{2} - 54 x)]_{1}^{e}

= [\frac{1}{18} (6 x^{3} ln (x) - 18 x^{2} ln (x) + 54 x ln (x) - 2 x^{3} + 9 x^{2} - 54 x)]_{1}^{e}

Compute the boundaries:

\frac{4 e ^{3} - 9 e ^{2}}{18} + \frac{47}{18}

= \frac{4 e ^{3} - 9 e ^{2}}{18} + \frac{47}{18}

Simplify

= \frac{4 e ^{3} - 9 e ^{2} + 47}{18}

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What is the integral from 1 to e of (x^2-2x+3)ln(x) ?
The integral from 1 to e of (x^2-2x+3)ln(x) is (4e^3-9e^2+47)/(18)