What is the integral of 3/5 (v-2)e^{2-4v+v^2} ?

The integral of 3/5 (v-2)e^{2-4v+v^2} is 3/10 e^{2-4v+v^2}+C

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integral of 3/5 (v-2)e^{2-4v+v^2}

\int \frac{3}{5} (v - 2) e^{2 - 4 v + v^{2}} d v

\frac{3}{10} e^{2 - 4 v + v^{2}} + C

Solution steps

\int \frac{3}{5} (v - 2) e^{2 - 4 v + v^{2}} d v

Take the constant out:

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

= \frac{3}{5} \cdot \int (v - 2) e^{2 - 4 v + v^{2}} d v

Apply u-substitution

: \int \frac{e ^{u}}{2} d u

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

Take the constant out:

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

= \frac{3}{5} \cdot \frac{1}{2} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{3}{5} \cdot \frac{1}{2} e^{u}

Substitute back

u = 2 - 4 v + v^{2}

= \frac{3}{5} \cdot \frac{1}{2} e^{2 - 4 v + v^{2}}

Simplify

\frac{3}{5} \cdot \frac{1}{2} e^{2 - 4 v + v^{2}} : \frac{3}{10} e^{2 - 4 v + v^{2}}

= \frac{3}{10} e^{2 - 4 v + v^{2}}

Add a constant to the solution

= \frac{3}{10} e^{2 - 4 v + v^{2}} + C

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What is the integral of 3/5 (v-2)e^{2-4v+v^2} ?
The integral of 3/5 (v-2)e^{2-4v+v^2} is 3/10 e^{2-4v+v^2}+C