# Integral of $ye^{-(x+1)y}$

Not sure where I'm going wrong on this one.

$$\int{ye^{-(x+1)y}}\:dy$$

$$u = y \qquad du = dy$$ $$dv = e^{-(x+1)y} \qquad v = -\frac{e^{-(x+1)y}}{x + 1}$$

$$-\frac{ye^{-(x+1)y}}{x + 1} \times \int{-\frac{e^{-(x+1)y}}{x + 1}}dy$$

Moving constants around

$$\frac{ye^{(x+1)y}}{(x + 1)^2} \times \int{e^{-(x+1)y}}\:dy$$

$$-\frac{ye^{2(x+1)y}}{(x + 1)^3}$$

So where did I make a mistake, as that doesn't match WolframAlpha, and I know that given $\int_{0}^{\infty}$ I should get $\frac{1}{(x+1)^2}$ which would indicate that y shouldn't be in the numerator, and I have an extra power in the denominator.

Thanks.

• It looks like you didn't apply integration by parts correctly. It's $\int f'g \, \mathrm dx = fg - \int f g' \, \mathrm dx$. Commented Nov 12, 2014 at 3:54
• @GFauxPas, I put my v, dv, u, du all in there. If I messed up there, please show how. Commented Nov 12, 2014 at 4:00
• @David, it is not multiplication (where you wrote $\;\times\;$ ), but substraction! Commented Nov 12, 2014 at 4:16

Looks like you're not applying integration by parts correctly.

$$-\frac{ye^{-(x+1)y}}{x + 1} \times \int{-\frac{e^{-(x+1)y}}{x + 1}}\,\mathrm dy$$

should be

$$-\frac{ye^{-(x+1)y}}{x + 1} - \int{-\frac{e^{-(x+1)y}}{x + 1}}\, \mathrm dy$$

it's:

$\int f'g \, \mathrm dx = fg - \int fg'\,\mathrm dx$

or

$\int v \, \mathrm du = uv - \int u \, \mathrm dv$

• Oops, thanks. I even looked at Wikipedia to make sure that I had the formula right. Much appreciated. Commented Nov 12, 2014 at 4:19
• Look up the proof for it, so you can rederive the formula for it if you ever forget it! Commented Nov 12, 2014 at 4:19