What is the integral of y(-4y-1) ?

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

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integral of y(-4y-1)

\int y (- 4 y - 1) d y

\frac{1}{2} y^{2} (- 4 y - 1) + \frac{2 y ^{3}}{3} + C

Solution steps

\int y (- 4 y - 1) d y

Apply Integration By Parts

: \frac{1}{2} y^{2} (- 4 y - 1) - \int - 2 y^{2} d y

= \frac{1}{2} y^{2} (- 4 y - 1) - \int - 2 y^{2} d y

\int - 2 y^{2} d y = - \frac{2 y ^{3}}{3}

= \frac{1}{2} y^{2} (- 4 y - 1) - (- \frac{2 y ^{3}}{3})

Simplify

= \frac{1}{2} y^{2} (- 4 y - 1) + \frac{2 y ^{3}}{3}

Add a constant to the solution

= \frac{1}{2} y^{2} (- 4 y - 1) + \frac{2 y ^{3}}{3} + C

f (x) = arcsin (3 x), at \frac{1}{6}

x \to \infty lim (cot (x))

\frac{d}{d t} (- 4.9 t^{2} + 24.5 t + 6.8)

f (x) = (9 x^{6} + 4 x^{3})^{4}

\int - \frac{1}{64} cos (8 x) d x

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