What is the integral from 0 to 1 of ln(4x) ?

The integral from 0 to 1 of ln(4x) is 2ln(2)-1

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integral from 0 to 1 of ln(4x)

\int_{0}^{1} ln (4 x) d x

2 ln (2) - 1

Solution steps

\int_{0}^{1} ln (4 x) d x

Apply Integration By Parts

: [x ln (4 x) - \int 1 d x]_{0}^{1}

= [x ln (4 x) - \int 1 d x]_{0}^{1}

\int 1 d x = x

= [x ln (4 x) - x]_{0}^{1}

Compute the boundaries:

2 ln (2) - 1

= 2 ln (2) - 1

\int - \frac{4}{x} d x

(x - 2) (x^{2} + 4)

6 cos (x)

- \frac{2}{9 ( 1 + x ) ^{\frac{5}{3}}}

\frac{d}{d x} (tan (\frac{π x}{2}))

What is the integral from 0 to 1 of ln(4x) ?
The integral from 0 to 1 of ln(4x) is 2ln(2)-1