What is the integral from 1 to 2 of x^2ln(x^4) ?

The integral from 1 to 2 of x^2ln(x^4) is -28/9+32/3 ln(2)

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integral from 1 to 2 of x^2ln(x^4)

\int_{1}^{2} x^{2} ln (x^{4}) d x

- \frac{28}{9} + \frac{32}{3} ln (2)

Decimal Notation

4.28245 \dots

Solution steps

\int_{1}^{2} x^{2} ln (x^{4}) d x

Apply Integration By Parts

: [\frac{1}{3} x^{3} ln (x^{4}) - \int \frac{4 x ^{2}}{3} d x]_{1}^{2}

= [\frac{1}{3} x^{3} ln (x^{4}) - \int \frac{4 x ^{2}}{3} d x]_{1}^{2}

\int \frac{4 x ^{2}}{3} d x = \frac{4 x ^{3}}{9}

= [\frac{1}{3} x^{3} ln (x^{4}) - \frac{4 x ^{3}}{9}]_{1}^{2}

\frac{1}{3} x^{3} ln (x^{4}) = \frac{4}{3} x^{3} ln (x)

= [\frac{4}{3} x^{3} ln (x) - \frac{4 x ^{3}}{9}]_{1}^{2}

Compute the boundaries:

- \frac{28}{9} + \frac{32}{3} ln (2)

= - \frac{28}{9} + \frac{32}{3} ln (2)

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What is the integral from 1 to 2 of x^2ln(x^4) ?
The integral from 1 to 2 of x^2ln(x^4) is -28/9+32/3 ln(2)