What is the integral from 1 to 36 of sqrt(t)ln(t) ?

The integral from 1 to 36 of sqrt(t)ln(t) is 288ln(6)-860/9

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integral from 1 to 36 of sqrt(t)ln(t)

\int_{1}^{36} t ln (t) d t

288 ln (6) - \frac{860}{9}

Decimal Notation

420.47117 \dots

Solution steps

\int_{1}^{36} t ln (t) d t

Apply Integration By Parts

: [\frac{2}{3} t^{\frac{3}{2}} ln (t) - \int \frac{2}{3} t^{\frac{1}{2}} d t]_{1}^{36}

= [\frac{2}{3} t^{\frac{3}{2}} ln (t) - \int \frac{2}{3} t^{\frac{1}{2}} d t]_{1}^{36}

\int \frac{2}{3} t^{\frac{1}{2}} d t = \frac{4}{9} t^{\frac{3}{2}}

= [\frac{2}{3} t^{\frac{3}{2}} ln (t) - \frac{4}{9} t^{\frac{3}{2}}]_{1}^{36}

Compute the boundaries:

288 ln (6) - \frac{860}{9}

= 288 ln (6) - \frac{860}{9}

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What is the integral from 1 to 36 of sqrt(t)ln(t) ?
The integral from 1 to 36 of sqrt(t)ln(t) is 288ln(6)-860/9