What is the integral from 1 to e of ln(6x) ?

The integral from 1 to e of ln(6x) is eln(6e)-e-ln(6)+1

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integral from 1 to e of ln(6x)

\int_{1}^{e} ln (6 x) d x

e ln (6 e) - e - ln (6) + 1

Solution steps

\int_{1}^{e} ln (6 x) d x

Apply Integration By Parts

: [x ln (6 x) - \int 1 d x]_{1}^{e}

= [x ln (6 x) - \int 1 d x]_{1}^{e}

\int 1 d x = x

= [x ln (6 x) - x]_{1}^{e}

Compute the boundaries:

e ln (6 e) - e - ln (6) + 1

= e ln (6 e) - e - ln (6) + 1

y = \frac{x ^{2} + 4}{3 x - x ^{2}}, (1, 2)

(arccot (x))^{5}

\int 3 sin (3 x) d x

\frac{d}{d x} (3 cos (x) - 5 cot (x))

(x y^{2}) d x + (1 + x^{2} y) d y = 0

What is the integral from 1 to e of ln(6x) ?
The integral from 1 to e of ln(6x) is eln(6e)-e-ln(6)+1