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

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

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

\int_{0}^{1} 2 ln (4 x) d x

2 (2 ln (2) - 1)

Solution steps

\int_{0}^{1} 2 ln (4 x) d x

Take the constant out:

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

= 2 \cdot \int_{0}^{1} ln (4 x) d x

Apply Integration By Parts

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

= 2 [x ln (4 x) - \int 1 d x]_{0}^{1}

\int 1 d x = x

= 2 [x ln (4 x) - x]_{0}^{1}

Compute the boundaries:

2 ln (2) - 1

= 2 (2 ln (2) - 1)

\int (2 - \frac{3}{x ^{10}}) d x

\int - 4 y^{3} d y

\int \frac{1}{x ^{2} - 4 x + 3} d x

\frac{d}{d y} (\frac{x ^{3} y - x y ^{3}}{x ^{2} + y ^{2}})

\int e^{x^{- \frac{1}{2}}} d x

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