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

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

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integral of e^{ty}(-4y-1)

\int e^{t y} (- 4 y - 1) d y

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

Solution steps

\int e^{t y} (- 4 y - 1) d y

Expand

(- 1 - 4 y) e^{t y} : - e^{t y} - 4 e^{t y} y

= \int - e^{t y} - 4 e^{t y} y d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int e^{t y} d y - \int 4 e^{t y} y d y

\int e^{t y} d y = \frac{1}{t} e^{t y}

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

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

Add a constant to the solution

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

(4 - \frac{1}{2 x ^{2}})^{- 1}

f (x) = x^{2} - 3 x - 4, (2, - 6)

\frac{d}{d x} (4 x^{3} + 4 x)

\frac{\partial}{\partial x} (\frac{x - 3 y}{x ^{2} + y ^{2}})

\int \frac{3 x - 1}{x ^{2} - x + 1} d x

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