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

The integral from 1 to 2 of x^4ln(x^2) is -62/25+64/5 ln(2)

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

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

- \frac{62}{25} + \frac{64}{5} ln (2)

Decimal

6.39228 \dots

Solution steps

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

Apply Integration By Parts

= [\frac{1}{5} x^{5} ln (x^{2}) - \int \frac{2 x ^{4}}{5} d x]_{1}^{2}

\int \frac{2 x ^{4}}{5} d x = \frac{2 x ^{5}}{25}

= [\frac{1}{5} x^{5} ln (x^{2}) - \frac{2 x ^{5}}{25}]_{1}^{2}

\frac{1}{5} x^{5} ln (x^{2}) = \frac{2}{5} x^{5} ln (x)

= [\frac{2}{5} x^{5} ln (x) - \frac{2 x ^{5}}{25}]_{1}^{2}

Compute the boundaries:

- \frac{62}{25} + \frac{64}{5} ln (2)

= - \frac{62}{25} + \frac{64}{5} ln (2)

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

\int_{0}^{1} \frac{1}{x ^{2} + 2} d x

\int_{- 1}^{1} 3 x + 1 d x

\int_{1}^{3} \frac{6}{9 + 9 x ^{2}} d x

\int_{x^{2}}^{3} sin^{4} (t) d t

What is the integral from 1 to 2 of x^4ln(x^2) ?
The integral from 1 to 2 of x^4ln(x^2) is -62/25+64/5 ln(2)