What is the integral from 1 to e of x^3*ln(x) ?

The integral from 1 to e of x^3*ln(x) is (3e^4+1)/(16)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to e of x^3*ln(x)

\int_{1}^{e} x^{3} \cdot ln (x) d x

\frac{3 e ^{4} + 1}{16}

Decimal Notation

10.29965 \dots

Solution steps

\int_{1}^{e} x^{3} ln (x) d x

Apply Integration By Parts

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

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

\int \frac{x ^{3}}{4} d x = \frac{x ^{4}}{16}

= [\frac{1}{4} (x^{4} ln (x) - 4 \cdot \frac{x ^{4}}{16})]_{1}^{e}

4 \cdot \frac{x ^{4}}{16} = \frac{x ^{4}}{4}

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

Compute the boundaries:

\frac{3 e ^{4}}{16} + \frac{1}{16}

= \frac{3 e ^{4}}{16} + \frac{1}{16}

Simplify

= \frac{3 e ^{4} + 1}{16}

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

y e^{- x} \frac{d y}{d x} = x

\frac{\partial}{\partial z} (x y + z^{3} x - 2 yz)

\int 18 cos^{2} (89 x) d x

arctan (t)

What is the integral from 1 to e of x^3*ln(x) ?
The integral from 1 to e of x^3*ln(x) is (3e^4+1)/(16)