What is the integral from 1 to 4 of-7sqrt(t)ln(t) ?

The integral from 1 to 4 of-7sqrt(t)ln(t) is -7(-28/9+32/3 ln(2))

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integral from 1 to 4 of-7sqrt(t)ln(t)

\int_{1}^{4} - 7 t ln (t) d t

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

Decimal Notation

- 29.97721 \dots

Solution steps

\int_{1}^{4} - 7 t ln (t) d t

Take the constant out:

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

= - 7 \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}

= - 7 [\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}}

= - 7 [\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)

= - 7 (- \frac{28}{9} + \frac{32}{3} ln (2))

\frac{d}{d x} (\frac{2 x ^{3} + 4}{x ^{2} - 4 x + 1})

x \to 0 lim (ln (tan (7 x)))

\int \frac{16 x ^{2}}{16 x ^{2} + 25} d x

\int \frac{x ^{2} + 4 x - 1}{x ^{3} - x} d x

f (x) = arctan (\frac{1 - x}{1 + x})

What is the integral from 1 to 4 of-7sqrt(t)ln(t) ?
The integral from 1 to 4 of-7sqrt(t)ln(t) is -7(-28/9+32/3 ln(2))