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

The integral from e^{-t} to e^{t-1 of}ln(x) is -2e^{t-1}+te^{t-1}+te^{-t}+e^{-t}

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

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

- 2 e^{t - 1} + t e^{t - 1} + t e^{- t} + e^{- t}

Solution steps

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

Apply Integration By Parts

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

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

\int 1 d x = x

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

Compute the boundaries:

- 2 e^{t - 1} + t e^{t - 1} + t e^{- t} + e^{- t}

= - 2 e^{t - 1} + t e^{t - 1} + t e^{- t} + e^{- t}

\int_{- π}^{π} cos (x) cos (3 x) d x

\int_{0}^{3} \frac{x}{2 + x ^{2}} d x

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

\int_{1}^{e^{2}} ln (2 x) d x

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

What is the integral from e^{-t} to e^{t-1 of}ln(x) ?
The integral from e^{-t} to e^{t-1 of}ln(x) is -2e^{t-1}+te^{t-1}+te^{-t}+e^{-t}