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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

\int 1 d t = t

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

Compute the boundaries:

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

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

\int F (x) d x

\frac{d}{d y} ((x^{2} + y^{2} + z^{2})^{- \frac{1}{2}} + ln (x y z))

\int tan^{7} (x) sec^{4} (x) d x

x \to - 2 lim (5 (3 x^{2} - 24 x + 3))

x \to 0 lim (\frac{- 9}{1 - 9 x})

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