What is the integral from 1 to e of ln(12x) ?

The integral from 1 to e of ln(12x) is eln(12e)-e-ln(12)+1

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integral from 1 to e of ln(12x)

\int_{1}^{e} ln (12 x) d x

e ln (12 e) - e - ln (12) + 1

Decimal

5.26976 \dots

Solution steps

\int_{1}^{e} ln (12 x) d x

Apply Integration By Parts

: [x ln (12 x) - \int 1 d x]_{1}^{e}

= [x ln (12 x) - \int 1 d x]_{1}^{e}

\int 1 d x = x

= [x ln (12 x) - x]_{1}^{e}

Compute the boundaries:

e ln (12 e) - e - ln (12) + 1

= e ln (12 e) - e - ln (12) + 1

y = x^{2}, y = x^{3}

x y^{^{'}} + y = - 8 x y^{2}

f (x) = x^{2} sin (x)

\int x + 8 d x

\frac{d}{d x} (x^{2} - 27)

What is the integral from 1 to e of ln(12x) ?
The integral from 1 to e of ln(12x) is eln(12e)-e-ln(12)+1