What is the integral from 1/7 to 5 of 16xln(7x) ?

The integral from 1/7 to 5 of 16xln(7x) is 16(-306/49+25/2 ln(35))

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integral from 1/7 to 5 of 16xln(7x)

\int_{\frac{1}{7}}^{5} 16 x ln (7 x) d x

16 (- \frac{306}{49} + \frac{25}{2} ln (35))

Decimal

611.15124 \dots

Solution steps

\int_{\frac{1}{7}}^{5} 16 x ln (7 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}{7}}^{5} x ln (7 x) d x

Apply Integration By Parts

: [\frac{1}{2} x^{2} ln (7 x) - \int \frac{x}{2} d x]_{\frac{1}{7}}^{5}

= 16 [\frac{1}{2} x^{2} ln (7 x) - \int \frac{x}{2} d x]_{\frac{1}{7}}^{5}

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

= 16 [\frac{1}{2} x^{2} ln (7 x) - \frac{x ^{2}}{4}]_{\frac{1}{7}}^{5}

Compute the boundaries:

- \frac{306}{49} + \frac{25}{2} ln (35)

= 16 (- \frac{306}{49} + \frac{25}{2} ln (35))

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

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

\frac{e ^{- 4 s}}{s ^{2} + 4 s + 3}

- 2 x + 3

y^{^{'}} = \frac{x - y}{2}

What is the integral from 1/7 to 5 of 16xln(7x) ?
The integral from 1/7 to 5 of 16xln(7x) is 16(-306/49+25/2 ln(35))