What is the integral from 1/3 to 5 of 16xln(3x) ?

The integral from 1/3 to 5 of 16xln(3x) is 16(-56/9+25/2 ln(15))

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integral from 1/3 to 5 of 16xln(3x)

\int_{\frac{1}{3}}^{5} 16 x ln (3 x) d x

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

Decimal Notation

442.05448 \dots

Solution steps

\int_{\frac{1}{3}}^{5} 16 x ln (3 x) d x

Take the constant out:

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

= 16 \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}

= 16 [\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}

= 16 [\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)

= 16 (- \frac{56}{9} + \frac{25}{2} ln (15))

f (x) = x^{2} + 4 x - 5, at x = 2

\frac{d}{d x} (n ln (x))

\int \frac{1}{25 - 81 x ^{2}} d x

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

\frac{d}{d x} ((\frac{x}{3} + 6 x)^{3})

What is the integral from 1/3 to 5 of 16xln(3x) ?
The integral from 1/3 to 5 of 16xln(3x) is 16(-56/9+25/2 ln(15))