What is the integral from 1/3 to 7 of 16xln(3x) ?

The integral from 1/3 to 7 of 16xln(3x) is 16(-110/9+49/2 ln(21))

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integral from 1/3 to 7 of 16xln(3x)

\int_{\frac{1}{3}}^{7} 16 x ln (3 x) d x

16 (- \frac{110}{9} + \frac{49}{2} ln (21))

Solution steps

\int_{\frac{1}{3}}^{7} 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}}^{7} 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}}^{7}

= 16 [\frac{1}{2} x^{2} ln (3 x) - \int \frac{x}{2} d x]_{\frac{1}{3}}^{7}

\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}}^{7}

Compute the boundaries:

- \frac{110}{9} + \frac{49}{2} ln (21)

= 16 (- \frac{110}{9} + \frac{49}{2} ln (21))

\frac{\partial}{\partial x} (arcsin (- 2 x^{4}))

n \to 0 lim (\frac{sin ( n )}{n})

\frac{d y}{d x} = \frac{y}{x} + \frac{y ^{2}}{x ^{2}}

\int (4 x + 3) cos (x) d x

y^{^{''}} + 3 y^{^{'}} + 2 = 0

What is the integral from 1/3 to 7 of 16xln(3x) ?
The integral from 1/3 to 7 of 16xln(3x) is 16(-110/9+49/2 ln(21))