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

Question

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.

Answer 1

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

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