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

The integral from 1 to e of ln(4x) is eln(4e)-e-2ln(2)+1

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integral from 1 to e of ln(4x)

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

e ln (4 e) - e - 2 ln (2) + 1

Decimal Notation

3.38204 \dots

Solution steps

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

Apply Integration By Parts

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

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

\int 1 d x = x

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

Compute the boundaries:

e ln (4 e) - e - 2 ln (2) + 1

= e ln (4 e) - e - 2 ln (2) + 1

(1 - x^{2}) \frac{d y}{d x} = x e^{y}

x \to + 5 lim (\frac{( x + 1 )}{x - 5})

x = 2 3 y, x = 0, y = 9

x \to - 2 lim ((x - 1)^{3})

2 x^{2}

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