What is the integral of 2ue^u ?

The integral of 2ue^u is 2(e^uu-e^u)+C

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integral of 2ue^u

\int 2 u e^{u} d u

2 (e^{u} u - e^{u}) + C

Solution steps

\int 2 u e^{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int u e^{u} d u

Apply Integration By Parts

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

= 2 (e^{u} u - \int e^{u} d u)

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

= 2 (e^{u} u - e^{u})

Add a constant to the solution

= 2 (e^{u} u - e^{u}) + C

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