Summer of 69!

For the sake of variety, I tried this:

Let $t = \ln(x + \sqrt{x^2 + 1})$

$\Rightarrow dt = \dfrac{1}{x + \sqrt{x^2 + 1}} \left(1 + \dfrac{x}{\sqrt{x^2 + 1}} \right) dx = \dfrac{1}{\sqrt{x^2 + 1}} dx$

$-t = -\ln(x + \sqrt{x^2 + 1}) = \ln(\sqrt{x^2 + 1} - x)$

$\Rightarrow e^t = x + \sqrt{x^2 + 1}$

$\Rightarrow e^{-t} = -x + \sqrt{x^2 + 1}$

$\Rightarrow \sqrt{x^2 + 1} = \dfrac{1}{2}(e^t + e^{-t}) \Rightarrow dt \dfrac{1}{2}(e^t + e^{-t}) = dx$

$I = \displaystyle \dfrac{1}{2} \int e^{69t} ( e^t + e^{-t} ) dt = \dfrac{1}{2}\int e^{70t} dt + \dfrac{1}{2} \int e^{68t} dt$

$I = \dfrac{1}{2} \left(\dfrac{e^{70t}}{70} + \dfrac{e^{68t}}{68} \right) + \alpha$

$I = \dfrac{1}{2} \left(\dfrac{(x+\sqrt{x^2 + 1})^{70}}{70} + \dfrac{(x+\sqrt{x^2 + 1})^{68}}{68} \right) + \alpha$

$a = 1 , b = 2 , c = 70 , d = 68 \Rightarrow a + b + c - d = \boxed{5}$

$I\quad =\quad \int { { (x+\sqrt { { x }^{ 2 }+1 } ) }^{ n } } dx\\ \\ let\quad x+\sqrt { { x }^{ 2 }+1 } \quad =\quad t\\ and\quad (1+\frac { x }{ \sqrt { { x }^{ 2 }+1 } } )dx\quad =\quad dt\\ \therefore \quad I\quad =\quad \frac { 1 }{ 2 } \int { { t }^{ n-2 } } ({ t }^{ 2 }+1)dt\\ On\quad solving,\\ I\quad =\quad \frac { 1 }{ 2 } \left[ \frac { { (x+\sqrt { { x }^{ 2 }+1 } ) }^{ n+1 } }{ n+1 } +\frac { { (x+\sqrt { { x }^{ 2 }+1 } ) }^{ n-1 } }{ n-1 } \right] +\alpha \\ on\quad substituting\quad values,\\ a+b+c-d\quad =\quad 1+2+70-68\quad =\quad 5$