What is the integral from 1 to 8 of sqrt(t)ln(t) ?

The integral from 1 to 8 of sqrt(t)ln(t) is 32sqrt(2)ln(2)+(-64sqrt(2)+4)/9

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integral from 1 to 8 of sqrt(t)ln(t)

\int_{1}^{8} t ln (t) d t

322 ln (2) + \frac{- 64 2 + 4}{9}

Decimal Notation

21.75607 \dots

Solution steps

\int_{1}^{8} 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}^{8}

= [\frac{2}{3} t^{\frac{3}{2}} ln (t) - \int \frac{2}{3} t^{\frac{1}{2}} d t]_{1}^{8}

\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}^{8}

Compute the boundaries:

322 ln (2) - \frac{64 2}{9} + \frac{4}{9}

= 322 ln (2) - \frac{64 2}{9} + \frac{4}{9}

Simplify

= 322 ln (2) + \frac{- 64 2 + 4}{9}

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What is the integral from 1 to 8 of sqrt(t)ln(t) ?
The integral from 1 to 8 of sqrt(t)ln(t) is 32sqrt(2)ln(2)+(-64sqrt(2)+4)/9