What is the integral from 1/7 to 5 of 10xln(7x) ?

The integral from 1/7 to 5 of 10xln(7x) is 10(-306/49+25/2 ln(35))

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integral from 1/7 to 5 of 10xln(7x)

\int_{\frac{1}{7}}^{5} 10 x ln (7 x) d x

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

Decimal Notation

381.96952 \dots

Solution steps

\int_{\frac{1}{7}}^{5} 10 x ln (7 x) d x

Take the constant out:

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

= 10 \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}

= 10 [\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}

= 10 [\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)

= 10 (- \frac{306}{49} + \frac{25}{2} ln (35))

\frac{\partial}{\partial x} (sin (\frac{y}{x + y}))

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

\frac{d y}{d x} + \frac{1}{x} y = 7 x^{2}

\frac{d}{d x} (\frac{6}{x ^{2} + 1})

\frac{d}{d t} (1 - 3 t)

What is the integral from 1/7 to 5 of 10xln(7x) ?
The integral from 1/7 to 5 of 10xln(7x) is 10(-306/49+25/2 ln(35))