What is the integral from 1 to 5 of x^3ln(4x) ?

The integral from 1 to 5 of x^3ln(4x) is -39-1/2 ln(2)+625/4 ln(20)

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integral from 1 to 5 of x^3ln(4x)

\int_{1}^{5} x^{3} ln (4 x) d x

- 39 - \frac{1}{2} ln (2) + \frac{625}{4} ln (20)

Decimal Notation

428.73659 \dots

Solution steps

\int_{1}^{5} x^{3} ln (4 x) d x

Apply Integration By Parts

: [\frac{1}{4} x^{4} ln (4 x) - \int \frac{x ^{3}}{4} d x]_{1}^{5}

= [\frac{1}{4} x^{4} ln (4 x) - \int \frac{x ^{3}}{4} d x]_{1}^{5}

\int \frac{x ^{3}}{4} d x = \frac{x ^{4}}{16}

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

Compute the boundaries:

- 39 - \frac{1}{2} ln (2) + \frac{625}{4} ln (20)

= - 39 - \frac{1}{2} ln (2) + \frac{625}{4} ln (20)

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

\frac{\partial}{\partial y} (x^{2} y + y z^{3} + z^{5} x)

\frac{d}{d x} ((x + \frac{1}{x}) (x - \frac{1}{x}))

y^{^{''}} + 10 y^{^{'}} + 25 = 0

\int (\frac{100}{25 + ( 5 y ) ^{2}} + e^{4}) d y

What is the integral from 1 to 5 of x^3ln(4x) ?
The integral from 1 to 5 of x^3ln(4x) is -39-1/2 ln(2)+625/4 ln(20)