What is the integral from 1 to 4 of sqrt(t)ln(t) ?

The integral from 1 to 4 of sqrt(t)ln(t) is -28/9+32/3 ln(2)

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integral from 1 to 4 of sqrt(t)ln(t)

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

- \frac{28}{9} + \frac{32}{3} ln (2)

Decimal Notation

4.28245 \dots

Solution steps

\int_{1}^{4} 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}^{4}

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

\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}^{4}

Compute the boundaries:

- \frac{28}{9} + \frac{32}{3} ln (2)

= - \frac{28}{9} + \frac{32}{3} ln (2)

y = x + 3 y = x + 1 y = - x + 1

\frac{d}{d x} (\frac{3 x + 2}{x - 1})

\frac{d}{d x} (\frac{e ^{x}}{x + 1})

\int \frac{e ^{\frac{x}{2}}}{2} d x

n = 0 \sum \infty (- \frac{5}{8})^{n}

What is the integral from 1 to 4 of sqrt(t)ln(t) ?
The integral from 1 to 4 of sqrt(t)ln(t) is -28/9+32/3 ln(2)