What is the integral of 2xe^x ?

The integral of 2xe^x is 2(e^xx-e^x)+C

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integral of 2xe^x

\int 2 x e^{x} d x

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

Solution steps

\int 2 x e^{x} d x

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

\frac{d}{d x} (ln (cos^{4} (x)))

\int (5 y^{4}) d y

\int (2 x - 1) y^{2} sin (4 x y) d x

f (x) = 8 x - 3

f (x) = 2 e^{x} cos (x), at x = 0