What is the integral from 1 to e of x^2ln(sqrt(x)) ?

The integral from 1 to e of x^2ln(sqrt(x)) is (2e^3+1)/(18)

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integral from 1 to e of x^2ln(sqrt(x))

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

\frac{2 e ^{3} + 1}{18}

Decimal Notation

2.28728 \dots

Solution steps

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

Simplify

ln (x) : \frac{1}{2} ln (x)

= \int_{1}^{e} \frac{1}{2} x^{2} ln (x) d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{1}^{e} x^{2} ln (x) d x

Apply Integration By Parts

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

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

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

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

3 \cdot \frac{x ^{3}}{9} = \frac{x ^{3}}{3}

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

Compute the boundaries:

\frac{2 e ^{3}}{9} + \frac{1}{9}

= \frac{1}{2} (\frac{2 e ^{3}}{9} + \frac{1}{9})

Simplify

= \frac{2 e ^{3} + 1}{18}

\frac{y ^{4}}{8} + \frac{1}{4 y ^{2}}

5 y^{^{''}} + 26 y = 0

e^{x^{2} + 9 x + 7}

n = 2 \sum \infty 2 n ln (n)

\frac{d}{d x} (\frac{x - 7}{x + 7})

What is the integral from 1 to e of x^2ln(sqrt(x)) ?
The integral from 1 to e of x^2ln(sqrt(x)) is (2e^3+1)/(18)