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

The integral from 1 to x^4 of t^2ln(t) is (4x^{12}ln(x))/3+(-x^{12}+1)/9

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

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

\frac{4 x ^{12} ln ( x )}{3} + \frac{- x ^{12} + 1}{9}

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{t ^{2}}{3} d t = \frac{t ^{3}}{9}

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

Compute the boundaries:

\frac{4}{3} x^{12} ln (x) - \frac{x ^{12}}{9} + \frac{1}{9}

= \frac{4}{3} x^{12} ln (x) - \frac{x ^{12}}{9} + \frac{1}{9}

Simplify

= \frac{4 x ^{12} ln ( x )}{3} + \frac{- x ^{12} + 1}{9}

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What is the integral from 1 to x^4 of t^2ln(t) ?
The integral from 1 to x^4 of t^2ln(t) is (4x^{12}ln(x))/3+(-x^{12}+1)/9