What is the integral from 1 to e^2 of x^3ln(x) ?

The integral from 1 to e^2 of x^3ln(x) is (7e^8+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^2 of x^3ln(x)

\int_{1}^{e^{2}} x^{3} ln (x) d x

\frac{7 e ^{8} + 1}{16}

Decimal Notation

1304.23161 \dots

Solution steps

\int_{1}^{e^{2}} 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^{2}}

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

\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^{2}}

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

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

Compute the boundaries:

\frac{7 e ^{8}}{16} + \frac{1}{16}

= \frac{7 e ^{8}}{16} + \frac{1}{16}

Simplify

= \frac{7 e ^{8} + 1}{16}

x + (\frac{1}{x})

f (x) = x^{2} - 4

\frac{d}{d x} (\frac{1}{2 - cos ( π ) x})

x \to - 6 + lim (\frac{1}{( x + 6 ) ^{4}})

- 2 y^{2} d y = y^{3} v d v

What is the integral from 1 to e^2 of x^3ln(x) ?
The integral from 1 to e^2 of x^3ln(x) is (7e^8+1)/(16)