There is another way to compute the symmetric BNE in increasing strategy.

Let $U(v)$ denote the expected utility of a player in equilibrium when his type is $v$: Given that the bidding strategy is increasing, a player with type $0$ will get the good with probability zero.

Thus he/she must bid zero and $U(0) = 0$ For any other $v > 0$, the probability that the player gets the good is $Q (v) = v^{n-1}$ (this is the probability that all the other players have a type lower than $v$) From the classes on mechanism design, we know that:

$U(v) = U(0) + \int_{0}^{v} Q(x) \,dx = \int_{0}^{v} x^{n-1} \,dx = \frac{v^{n}}{n}$

On the other hand, we can write $U(v)$ as

$U(v)= (v-\alpha b(v)))v^{n-1}$

Therefore

$(v-\alpha b(v))v^{n-1} = \frac{v^{n}}{n}$

and

$b(v) = \frac{n-1}{n \alpha} v$

There is another way to compute the symmetric BNE in increasing strategy.

Let $U(v)$ denote the expected utility of a player in equilibrium when his type is $v$: Given that the bidding strategy is increasing, a player with type $0$ will get the good with probability zero.

Thus he/she must bid zero and $U(0) = 0$. For any other $v > 0$, the probability that the player gets the good is $Q (v) = v^{n-1}$ (this is the probability that all the other players have a type lower than $v$) From the classes on mechanism design, we know that:

$U(v) = U(0) + \int_{0}^{v} Q(x) \,dx = \int_{0}^{v} x^{n-1} \,dx = \frac{v^{n}}{n}$

On the other hand, we can write $U(v)$ as

$U(v)= (v-\alpha b(v)))v^{n-1}$

Therefore

$(v-\alpha b(v))v^{n-1} = \frac{v^{n}}{n}$

and

$b(v) = \frac{n-1}{n \alpha} v$