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

The integral from 1 to e^x of ln(x) is e^xx-e^x+1

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integral from 1 to e^x of ln(x)

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

e^{x} x - e^{x} + 1

Solution steps

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

Apply Integration By Parts

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

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

\int 1 d x = x

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

Compute the boundaries:

e^{x} x - e^{x} + 1

= e^{x} x - e^{x} + 1

\int \frac{9 x ^{2} + 7 x + 9}{( x ^{2} + 1 ) ^{2}} d x

\int \frac{5 - e ^{x}}{e ^{4 x}} d x

\frac{\partial}{\partial x} (ln (x + 9 y + 5 z))

\frac{d}{d x} (4 a e^{4 x})

\int_{- \infty}^{2} x^{2} e^{x} d x

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