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

The integral from 1 to e of (1+x^2)ln(x) is (2e^3+10)/9

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

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

\frac{2 e ^{3} + 10}{9}

Decimal

5.57456 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{2 e ^{3}}{9} + \frac{10}{9}

= \frac{2 e ^{3}}{9} + \frac{10}{9}

Simplify

= \frac{2 e ^{3} + 10}{9}

What is the integral from 1 to e of (1+x^2)ln(x) ?
The integral from 1 to e of (1+x^2)ln(x) is (2e^3+10)/9