# Confused with integral and natural logarithm When reading about ideal gas and adiabatic expansion, I got stuck with the following: W_(ab)= int_(V_a)^(V_b)(dV)/(V^alpha)=(V_b^(1- alpha)-V_a^(1- alpha))/(1-alpha)

Confused with integral and natural logarithm
${W}_{ab}={\int }_{{\mathit{V}}_{\mathit{a}}}^{{\mathit{V}}_{\mathit{b}}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{dV}{{V}^{\alpha }}=\frac{{V}_{b}^{1-\alpha }-{V}_{a}^{1-\alpha }}{1-\alpha }$
I know the following rule, but I couldn't come to the above:
${\int }_{{\mathit{V}}_{\mathit{a}}}^{{\mathit{V}}_{\mathit{b}}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{dV}{V}=ln|\frac{{V}_{b}}{{V}_{a}}|$
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inmholtau5
You can write it as

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.

b is a unknown factor.
If you differentiate $b\cdot {V}^{1-\alpha }$ you get $b\cdot \left(1-\alpha \right){V}^{-\alpha }$
Thus the factor b has to be $\frac{1}{1-\alpha }$