What is the integral of ye^{(t-1)y} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of ye^{(t-1)y}

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

Add a constant to the solution

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

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