*Pukes Mathematically II

$I=\int _{ 0 }^{ e }{ \sqrt { 1-\ln { \left( x \right) } } dx }$

$let\quad u=1-\ln { \left( x \right) } \Rightarrow x={ e }^{ \left( 1-u \right) }\Rightarrow dx=-{ e }^{ \left( 1-u \right) }du$

$I=\int _{ 0 }^{ \infty }{ { u }^{ \frac { 1 }{ 2 } }{ e }^{ \left( 1-u \right) }du } =e\int _{ 0 }^{ \infty }{ { u }^{ \frac { 1 }{ 2 } }{ e }^{ -u }du }$

$I=e\cdot \Gamma \left( \frac { 3 }{ 2 } \right) =e\cdot \frac { 1 }{ 2 } \cdot \Gamma \left( \frac { 1 }{ 2 } \right) \qquad$ Gamma Function

$I=\frac { 1 }{ 2 } e\sqrt { \pi }$

$A=\frac { 1 }{ 2 } ,B=1,C=\frac { 1 }{ 2 }$

$A+B+C=2$

Let $f(x)=\sqrt{1-\ln(x)}$

First, we notice that $f(x)$ is strictly decreasing in its domain. Therefore, $f(x)$ has an inverse function $f^{-1}(x)$ , which can be easily found to be $f^{-1}(x)=e^{1-x^{2}}$

If we look at our integral in the plane, we see that

$I = \int_{0}^{e} f(x) \ dx = -\int_{f(0)}^{f(e)} f^{-1}(y) \ dy = \int_{0}^{\infty} e^{1-y^{2}} \ dy$

We can rewrite the last form for $I$ as follows;

$I = e \int_{0}^{\infty} e^{-y^{2}}dy$

Which contains the well known Gaussian Integral. Therefore, $I = \frac{e \sqrt{\pi}}{2}$ and we can obtain our final answer of $\frac{1}{2}+1+\frac{1}{2}=\boxed{2}$