What is the integral from 1 to 4 of 6sqrt(t)ln(t) ?

The integral from 1 to 4 of 6sqrt(t)ln(t) is 6(-28/9+32/3 ln(2))

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integral from 1 to 4 of 6sqrt(t)ln(t)

\int_{1}^{4} 6 t ln (t) d t

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

Decimal Notation

25.69475 \dots

Solution steps

\int_{1}^{4} 6 t ln (t) d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \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}

= 6 [\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}}

= 6 [\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)

= 6 (- \frac{28}{9} + \frac{32}{3} ln (2))

\int sec (2 t) tan (2 t) d t

(8 x - 6)^{2}

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\int (\frac{x ^{2} + 5}{( x + 2 ) ^{2}}) d x

\int \frac{x ^{3}}{x + 2} d x

What is the integral from 1 to 4 of 6sqrt(t)ln(t) ?
The integral from 1 to 4 of 6sqrt(t)ln(t) is 6(-28/9+32/3 ln(2))