What is the integral of 3(8y-1)e^{4y^2-y} ?

The integral of 3(8y-1)e^{4y^2-y} is 3e^{4y^2-y}+C

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integral of 3(8y-1)e^{4y^2-y}

\int 3 (8 y - 1) e^{4 y^{2} - y} d y

3 e^{4 y^{2} - y} + C

Solution steps

\int 3 (8 y - 1) e^{4 y^{2} - y} d y

Take the constant out:

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

= 3 \cdot \int (8 y - 1) e^{4 y^{2} - y} d y

Apply u-substitution

= 3 \cdot \int e^{u} d u

Use the common integral:

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

= 3 e^{u}

Substitute back

u = 4 y^{2} - y

= 3 e^{4 y^{2} - y}

Add a constant to the solution

= 3 e^{4 y^{2} - y} + C

What is the integral of 3(8y-1)e^{4y^2-y} ?
The integral of 3(8y-1)e^{4y^2-y} is 3e^{4y^2-y}+C