What is the integral from 1 to e of ln(19x) ?

The integral from 1 to e of ln(19x) is eln(19e)-e-ln(19)+1

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integral from 1 to e of ln(19x)

\int_{1}^{e} ln (19 x) d x

e ln (19 e) - e - ln (19) + 1

Decimal Notation

6.05937 \dots

Solution steps

\int_{1}^{e} ln (19 x) d x

Apply Integration By Parts

: [x ln (19 x) - \int 1 d x]_{1}^{e}

= [x ln (19 x) - \int 1 d x]_{1}^{e}

\int 1 d x = x

= [x ln (19 x) - x]_{1}^{e}

Compute the boundaries:

e ln (19 e) - e - ln (19) + 1

= e ln (19 e) - e - ln (19) + 1

x \to 8 + lim (\frac{1}{x + 8})

\frac{\partial}{\partial x} (x ln (z^{2} + x^{2}))

\frac{d}{d x} ((x^{2} + 2) (x^{4} + 1))

\frac{\partial}{\partial x} (9 x e^{x})

\int \frac{1}{4 ( x + 1 )} d x

What is the integral from 1 to e of ln(19x) ?
The integral from 1 to e of ln(19x) is eln(19e)-e-ln(19)+1