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

The integral from 1 to e of t^2ln(t) is (2e^3+1)/9

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integral from 1 to e of t^2ln(t)

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

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

Decimal Notation

4.57456 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

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

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

Compute the boundaries:

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

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

Simplify

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

\frac{x - 3}{x}

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

x^{3} y^{^{'}} + 2 x^{2} y = 4 cos (4 x) y^{- \frac{1}{2}}

f (x) = lo g_{a} (x)

\frac{\partial}{\partial x} (x^{\frac{1}{2}} + y)

What is the integral from 1 to e of t^2ln(t) ?
The integral from 1 to e of t^2ln(t) is (2e^3+1)/9