What is the integral of ue^{-u} ?

The integral of ue^{-u} is -e^{-u}u-e^{-u}+C

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integral of ue^{-u}

\int u e^{- u} d u

- e^{- u} u - e^{- u} + C

Solution steps

\int u e^{- u} d u

Apply Integration By Parts

: - e^{- u} u - \int - e^{- u} d u

= - e^{- u} u - \int - e^{- u} d u

\int - e^{- u} d u = e^{- u}

= - e^{- u} u - e^{- u}

Add a constant to the solution

= - e^{- u} u - e^{- u} + C

y = 4 ln (x), at x = 4

\int_{0}^{1} (x^{\frac{4}{5}}) d x

\int \frac{( 1 + 6 ln ( x ) ) ^{11}}{x} d x

\int x^{2} + 2 x y + y^{2} d x

\frac{d y}{d x} = \frac{1}{1 - x ^{2}} - \frac{1}{x}