What substitution should I make?

Assume $x+\sqrt{x^2+1}=z\Rightarrow \dfrac{1}{z}=\sqrt{x^2+1}-x$ .
Differentiating we get that, $\dfrac{dz}{dx}=1+\dfrac{2x}{2\sqrt{x^2+1}}$ .
Note that $z+\dfrac{1}{z}=2\sqrt{x^2+1} \hspace{.3cm} \text{ and } \hspace{.3cm} z-\dfrac{1}{z}=2x$

Therefore, $\dfrac{dz}{dx}=1+\dfrac{z-\frac{1}{z}}{z+\frac{1}{z}}=1+\dfrac{z^2-1}{z^2+1}=\dfrac{2z^2}{z^2+1}$ Now, after changing the limits, the integral becomes $\int_{1}^{1+\sqrt{2}} \sqrt{z}\dfrac{z^2+1}{2z^2}dz$ Now, use the substitution $z=y^2$ and the the integral becomes, $\int_{1}^{\sqrt{1+\sqrt{2}}}2y^2\dfrac{(y^4+1)}{2y^4}dy \\ =\int_{1}^{\sqrt{1+\sqrt{2}}}\dfrac{(y^4+1)}{y^2}dy$ The above integral is a trivial one from there!

$we\quad have\quad the\quad integral\quad I\quad =\int _{ 0 }^{ 1 }{ \sqrt { x+\sqrt { { x }^{ 2 }+1 } } } dx\\ remember\quad the\quad function:\quad \sinh ^{ -1 }{ (x)=\ln { (x+\sqrt { { x }^{ 2 }+1 } ) } } \\ \therefore \quad \frac { 1 }{ 2 } \sinh ^{ -1 }{ x } =\ln { \sqrt { x+\sqrt { { x }^{ 2 }+1 } } } \\ \therefore \quad { e }^{ \frac { \sinh ^{ -1 }{ x } }{ 2 } }=\sqrt { x+\sqrt { { x }^{ 2 }+1 } } \\ \therefore \quad I\quad =\quad \int _{ 0 }^{ 1 }{ { e }^{ \frac { \sinh ^{ -1 }{ x } }{ 2 } } } dx\\ now\quad put\quad :\quad x\quad =\quad \sinh { 2u } \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad when\quad x\quad =\quad 0\quad \quad \quad \quad u\quad =\quad 0\\ \quad \quad \quad \quad \quad \quad \quad \therefore \quad dx\quad =\quad 2\cosh { 2u } du\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad x\quad =\quad 1\quad \quad \quad \quad \quad \quad u\quad =\quad \ln { \sqrt { 1+\sqrt { 2 } } } \\ \therefore \quad I\quad =\quad 2\int _{ 0 }^{ \ln { \sqrt { 1+\sqrt { 2 } } } }{ { e }^{ u } } \cosh { 2u } \quad du\\ and\quad we\quad have\quad \cosh { 2u } =\frac { { e }^{ 2u }+{ e }^{ -2u } }{ 2 } \\ by\quad sub\quad in\quad I\quad =\quad \int _{ 0 }^{ \ln { \sqrt { 1+\sqrt { 2 } } } }{ { e }^{ 3u } } +{ e }^{ -u }\quad du\quad =\frac { 1 }{ 3{ e }^{ u } } ({ e }^{ 4u }-3)+c\quad \\ by\quad solving\quad we\quad get\quad I\quad =\quad \frac { 2 }{ 3 } (\sqrt { \sqrt { 8 } -2 } +1)\quad so\quad A+B+C+D=2+3+8+2=15\quad \\$