What is the integral from 1/3 to 5 of 8xln(3x) ?

The integral from 1/3 to 5 of 8xln(3x) is 8(-56/9+25/2 ln(15))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1/3 to 5 of 8xln(3x)

\int_{\frac{1}{3}}^{5} 8 x ln (3 x) d x

8 (- \frac{56}{9} + \frac{25}{2} ln (15))

Decimal Notation

221.02724 \dots

Solution steps

\int_{\frac{1}{3}}^{5} 8 x ln (3 x) d x

Take the constant out:

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

= 8 \cdot \int_{\frac{1}{3}}^{5} x ln (3 x) d x

Apply Integration By Parts

: [\frac{1}{2} x^{2} ln (3 x) - \int \frac{x}{2} d x]_{\frac{1}{3}}^{5}

= 8 [\frac{1}{2} x^{2} ln (3 x) - \int \frac{x}{2} d x]_{\frac{1}{3}}^{5}

\int \frac{x}{2} d x = \frac{x ^{2}}{4}

= 8 [\frac{1}{2} x^{2} ln (3 x) - \frac{x ^{2}}{4}]_{\frac{1}{3}}^{5}

Compute the boundaries:

- \frac{56}{9} + \frac{25}{2} ln (15)

= 8 (- \frac{56}{9} + \frac{25}{2} ln (15))

\frac{d}{d x} (ln (sin (x^{2})))

\frac{d}{d x} (\frac{1}{f ( x ) ^{2} ( x )})

- 10 x^{3} + 6 x^{40} y + y^{4} = 3

\int \frac{x ^{3}}{1 - x ^{2}} d x

x \to 4 - lim (2 - x - x^{2})

What is the integral from 1/3 to 5 of 8xln(3x) ?
The integral from 1/3 to 5 of 8xln(3x) is 8(-56/9+25/2 ln(15))