What is the integral from 1 to 5 of sqrt(t)ln(t) ?

The integral from 1 to 5 of sqrt(t)ln(t) is (10sqrt(5))/3 ln(5)+(-20sqrt(5)+4)/9

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integral from 1 to 5 of sqrt(t)ln(t)

\int_{1}^{5} t ln (t) d t

\frac{10 5}{3} ln (5) + \frac{- 20 5 + 4}{9}

Decimal Notation

7.47144 \dots

Solution steps

\int_{1}^{5} 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}^{5}

= [\frac{2}{3} t^{\frac{3}{2}} ln (t) - \int \frac{2}{3} t^{\frac{1}{2}} d t]_{1}^{5}

\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}^{5}

Compute the boundaries:

\frac{10 5}{3} ln (5) - \frac{20 5}{9} + \frac{4}{9}

= \frac{10 5}{3} ln (5) - \frac{20 5}{9} + \frac{4}{9}

Simplify

= \frac{10 5}{3} ln (5) + \frac{- 20 5 + 4}{9}

y^{^{'}} = sec^{2} (x) (2 + y)^{2}, y (π) = - 5

y^{^{''}} + 4 y = 3 sin (2 x)

2^{x^{2} + 1}

\frac{9}{x + 2}

What is the integral from 1 to 5 of sqrt(t)ln(t) ?
The integral from 1 to 5 of sqrt(t)ln(t) is (10sqrt(5))/3 ln(5)+(-20sqrt(5)+4)/9