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

The integral of y(2y+1)e^{4y^3+3y^2-4} is 1/6 e^{4y^3+3y^2-4}+C

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

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

\frac{1}{6} e^{4 y^{3} + 3 y^{2} - 4} + C

Solution steps

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

Apply u-substitution

: \int \frac{e ^{u}}{6} d u

= \int \frac{e ^{u}}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{6} e^{u}

Substitute back

u = 4 y^{3} + 3 y^{2} - 4

= \frac{1}{6} e^{4 y^{3} + 3 y^{2} - 4}

Add a constant to the solution

= \frac{1}{6} e^{4 y^{3} + 3 y^{2} - 4} + C

5 x^{2} - x^{3}

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

x \frac{d y}{d x} + 2 y = 5 x^{3}

\int \frac{39}{39 + e ^{x}} d x

x \to - 2 lim (5 x + 2)

What is the integral of y(2y+1)e^{4y^3+3y^2-4} ?
The integral of y(2y+1)e^{4y^3+3y^2-4} is 1/6 e^{4y^3+3y^2-4}+C