What is the integral from 1 to 3 of (2-x)^2ln(4x) ?

The integral from 1 to 3 of (2-x)^2ln(4x) is 3ln(12)-26/9-14/3 ln(2)

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integral from 1 to 3 of (2-x)^2ln(4x)

\int_{1}^{3} (2 - x)^{2} ln (4 x) d x

3 ln (12) - \frac{26}{9} - \frac{14}{3} ln (2)

Decimal

1.33114 \dots

Solution steps

\int_{1}^{3} (2 - x)^{2} ln (4 x) d x

Apply Integration By Parts

: [ln (4 x) (4 x - 2 x^{2} + \frac{x ^{3}}{3}) - \int \frac{1}{3} (12 - 6 x + x^{2}) d x]_{1}^{3}

= [ln (4 x) (4 x - 2 x^{2} + \frac{x ^{3}}{3}) - \int \frac{1}{3} (12 - 6 x + x^{2}) d x]_{1}^{3}

\int \frac{1}{3} (12 - 6 x + x^{2}) d x = 4 x - x^{2} + \frac{x ^{3}}{9}

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

Simplify

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

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

Compute the boundaries:

3 ln (12) - \frac{26}{9} - \frac{14}{3} ln (2)

= 3 ln (12) - \frac{26}{9} - \frac{14}{3} ln (2)

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What is the integral from 1 to 3 of (2-x)^2ln(4x) ?
The integral from 1 to 3 of (2-x)^2ln(4x) is 3ln(12)-26/9-14/3 ln(2)