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

The integral from 1 to 7 of sqrt(t)ln(t) is (14sqrt(7))/3 ln(7)+(-28sqrt(7)+4)/9

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

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

\frac{14 7}{3} ln (7) + \frac{- 28 7 + 4}{9}

Solution steps

\int_{1}^{7} 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}^{7}

= [\frac{2}{3} t^{\frac{3}{2}} ln (t) - \int \frac{2}{3} t^{\frac{1}{2}} d t]_{1}^{7}

\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}^{7}

Compute the boundaries:

\frac{14 7}{3} ln (7) - \frac{28 7}{9} + \frac{4}{9}

= \frac{14 7}{3} ln (7) - \frac{28 7}{9} + \frac{4}{9}

Simplify

= \frac{14 7}{3} ln (7) + \frac{- 28 7 + 4}{9}

\frac{3}{s ( s ^{2} + s + 5 )}

e^{t} sin (2 t)

\int_{1}^{2} x 2^{(x^{2})} d x

\int \frac{( x )}{1 - x ^{2}} d x

\int_{0}^{\infty} x^{- 3} d x

What is the integral from 1 to 7 of sqrt(t)ln(t) ?
The integral from 1 to 7 of sqrt(t)ln(t) is (14sqrt(7))/3 ln(7)+(-28sqrt(7)+4)/9