What is the integral of 4xe^x ?

The integral of 4xe^x is 4(e^xx-e^x)+C

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integral of 4xe^x

\int 4 x e^{x} d x

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

Solution steps

\int 4 x e^{x} d x

Take the constant out:

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

= 4 \cdot \int x e^{x} d x

Apply Integration By Parts

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

= 4 (e^{x} x - \int e^{x} d x)

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

= 4 (e^{x} x - e^{x})

Add a constant to the solution

= 4 (e^{x} x - e^{x}) + C

t \to 0 + lim (t + 1)

\int \frac{x ^{2}}{cos ^{2} ( x )} d x

\int 3 v (v^{2} + 6)^{2} d v

\frac{\partial}{\partial x} (4 x^{3} y^{3} - 4)

x^{3} d y + x y d x = x^{2} d y + 2 y d x