What is the integral from 1/3 to 5 of 6xln(3x) ?

The integral from 1/3 to 5 of 6xln(3x) is 6(-56/9+25/2 ln(15))

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integral from 1/3 to 5 of 6xln(3x)

\int_{\frac{1}{3}}^{5} 6 x ln (3 x) d x

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

Decimal Notation

165.77043 \dots

Solution steps

\int_{\frac{1}{3}}^{5} 6 x ln (3 x) d x

Take the constant out:

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

= 6 \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}

= 6 [\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}

= 6 [\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)

= 6 (- \frac{56}{9} + \frac{25}{2} ln (15))

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

ln (x), at 1

\frac{d}{d x} ((4 x + 1) (1 - x)^{3})

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

x \to \infty lim (\frac{π}{3})

What is the integral from 1/3 to 5 of 6xln(3x) ?
The integral from 1/3 to 5 of 6xln(3x) is 6(-56/9+25/2 ln(15))