When reading about ideal gas and adiabatic expansion, I got stuck with the following: $$W_{ab}=\int_{{\it V_a}}^{{\it V_b}}\!\,{\rm }\frac{dV}{V^\alpha}=\frac{V_b^{1-\alpha}-V_a^{1-\alpha}}{1-\alpha}$$
I know the following rule, but I couldn't come to the above: $$\int_{{\it V_a}}^{{\it V_b}}\!\,{\rm }\frac{dV}{V}=ln|\frac{V_b}{V_a}|$$
You can write it as $$\int_{V_a}^{V_b} V^{-\alpha} \ dV$$ It is obvious, that a 1 has to be added to the exponent to get the integral. If you differentiate a polynomial, the exponent decreases by 1.
$$\int_{V_a}^{V_b} V^{-\alpha} \ dV= b\cdot V^{1-\alpha}\bigg|_{V_a}^{V_b}$$
b is a unknown factor.
If you differentiate $b\cdot V^{1-\alpha}$ you get $b\cdot (1-\alpha ) V^{-\alpha}$.
Thus the factor b has to be $\frac{1}{1-\alpha}$.