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

The integral from 1 to 4 of sqrt(x)ln(sqrt(x)) is -14/9+16/3 ln(2)

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

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

- \frac{14}{9} + \frac{16}{3} ln (2)

Decimal Notation

2.14122 \dots

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{1}{3} x^{\frac{1}{2}} d x = \frac{2}{9} x^{\frac{3}{2}}

= [\frac{2}{3} x^{\frac{3}{2}} ln (x) - \frac{2}{9} x^{\frac{3}{2}}]_{1}^{4}

Simplify

[\frac{2}{3} x^{\frac{3}{2}} ln (x) - \frac{2}{9} x^{\frac{3}{2}}]_{1}^{4} : [\frac{1}{3} x^{\frac{3}{2}} ln (x) - \frac{2}{9} x^{\frac{3}{2}}]_{1}^{4}

= [\frac{1}{3} x^{\frac{3}{2}} ln (x) - \frac{2}{9} x^{\frac{3}{2}}]_{1}^{4}

Compute the boundaries:

- \frac{14}{9} + \frac{16}{3} ln (2)

= - \frac{14}{9} + \frac{16}{3} ln (2)

\int \frac{x}{5 - 4 x ^{2} - x ^{4}} d x

\frac{12 e ^{- s}}{s ( 10 s + 1 )}

\int_{0}^{8} \frac{3}{8 - x} d x

\int \frac{2}{x ^{3} + 9 x} d x

\frac{d}{d x} ((tan (x))^{s i n (x)})

What is the integral from 1 to 4 of sqrt(x)ln(sqrt(x)) ?
The integral from 1 to 4 of sqrt(x)ln(sqrt(x)) is -14/9+16/3 ln(2)