What is the integral of 2/3 xe^{3/(-5)} ?

The integral of 2/3 xe^{3/(-5)} is (x^2)/(3e^{3/5)}+C

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integral of 2/3 xe^{3/(-5)}

\int \frac{2}{3} x e^{\frac{3}{- 5}} d x

\frac{x ^{2}}{3 e ^{\frac{3}{5}}} + C

Solution steps

\int \frac{2}{3} x e^{\frac{3}{- 5}} d x

Take the constant out:

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

= \frac{2}{3} e^{\frac{3}{- 5}} \cdot \int x d x

\frac{2}{3} e^{\frac{3}{- 5}} = \frac{2}{3 e ^{\frac{3}{5}}}

= \frac{2}{3 e ^{\frac{3}{5}}} \cdot \int x d x

Apply the Power Rule

= \frac{2}{3 e ^{\frac{3}{5}}} \cdot \frac{x ^{2}}{2}

Simplify

\frac{2}{3 e ^{\frac{3}{5}}} \cdot \frac{x ^{2}}{2} : \frac{x ^{2}}{3 e ^{\frac{3}{5}}}

= \frac{x ^{2}}{3 e ^{\frac{3}{5}}}

Add a constant to the solution

= \frac{x ^{2}}{3 e ^{\frac{3}{5}}} + C

What is the integral of 2/3 xe^{3/(-5)} ?
The integral of 2/3 xe^{3/(-5)} is (x^2)/(3e^{3/5)}+C