What is the integral from 1 to e of x^5ln(x^2) ?

The integral from 1 to e of x^5ln(x^2) is (5e^6+1)/(18)

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integral from 1 to e of x^5ln(x^2)

\int_{1}^{e} x^{5} ln (x^{2}) d x

\frac{5 e ^{6} + 1}{18}

Decimal Notation

112.11910 \dots

Solution steps

\int_{1}^{e} x^{5} ln (x^{2}) d x

Apply Integration By Parts

: [\frac{1}{6} (x^{6} ln (x^{2}) - 6 \cdot \int \frac{x ^{5}}{3} d x)]_{1}^{e}

= [\frac{1}{6} (x^{6} ln (x^{2}) - 6 \cdot \int \frac{x ^{5}}{3} d x)]_{1}^{e}

\int \frac{x ^{5}}{3} d x = \frac{x ^{6}}{18}

= [\frac{1}{6} (x^{6} ln (x^{2}) - 6 \cdot \frac{x ^{6}}{18})]_{1}^{e}

Simplify

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

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

Compute the boundaries:

\frac{5 e ^{6}}{18} + \frac{1}{18}

= \frac{5 e ^{6}}{18} + \frac{1}{18}

Simplify

= \frac{5 e ^{6} + 1}{18}

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What is the integral from 1 to e of x^5ln(x^2) ?
The integral from 1 to e of x^5ln(x^2) is (5e^6+1)/(18)