What is the integral of e^{1/3 x^2-2x}(4/3 x-4) ?

The integral of e^{1/3 x^2-2x}(4/3 x-4) is 2e^{1/3 x^2-2x}+C

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integral of e^{1/3 x^2-2x}(4/3 x-4)

\int e^{\frac{1}{3} x^{2} - 2 x} (\frac{4}{3} x - 4) d x

2 e^{\frac{1}{3} x^{2} - 2 x} + C

Solution steps

\int e^{\frac{1}{3} x^{2} - 2 x} (\frac{4}{3} x - 4) d x

Apply u-substitution

: \int 2 e^{u} d u

= \int 2 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 e^{u} d u

Use the common integral:

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

= 2 e^{u}

Substitute back

u = \frac{1}{3} x^{2} - 2 x

= 2 e^{\frac{1}{3} x^{2} - 2 x}

Add a constant to the solution

= 2 e^{\frac{1}{3} x^{2} - 2 x} + C

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What is the integral of e^{1/3 x^2-2x}(4/3 x-4) ?
The integral of e^{1/3 x^2-2x}(4/3 x-4) is 2e^{1/3 x^2-2x}+C