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

Popular >

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

\int e^{2 x} \cdot sin (x) d x

f (x) = \frac{1}{6} x^{4}, at x = - 1

1 - e^{- t} sin (2 t)

\frac{d}{d x} (8 x (x^{2} + 3)^{3})

f (x) = 16 - x^{2}

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