JEE Advanced Binomial Theorem 1

Set $a=0$ , then given expression becomes:

${ x }^{ 100 }=\left( \begin{matrix} 100 \\ 0 \end{matrix} \right) { x }^{ 100 }$

Substituting in integral, we get:

$\int { \frac { 2x }{ { x }^{ 200 } }dx }$

Now put $x^{2}=t$

$\Rightarrow \int { \frac { 2x }{ { x }^{ 200 } } dx } =\int { \frac { dt }{ { t }^{ 100 } } } =\frac { -1 }{ { 99t }^{ 99 } }+C$

Now put back $x^{2}$ in place of $t$

$=\frac { -1 }{ { 99({ x }^{ 2 }) }^{ 99 } }+C$

The answer is: $99+99+2-1=\boxed{199}$

Classic JEE style. But I would like to know the actual solution, @Pranjal Jain

Solution inspired from Ronak Agarwal's comment on Raghav Vaidyanathan's solution

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Assume that $a = \sqrt{-1} = i$ (assumption is valid as the integration is not done w.r.t. $a$ ).

Thus, our binomial expression becomes

$\begin{aligned} {(x+i)}^{100} &= \underbrace{\dbinom{100}{0} x^{100}}_{t_0} + \underbrace{\dbinom{100}{1} x^{99} i}_{t_1} - \underbrace{\dbinom{100}{2} x^{98}}_{(- t_2)} - \underbrace{\dbinom{100}{3} x^{97} i}_{(- t_3)} + \underbrace{\dbinom{100}{4} x^{96}}_{t_4} + \cdots - \underbrace{\dbinom{100}{98} x^2}_{(- t_{98})} - \underbrace{\dbinom{100}{99} x i}_{(- t_{99})} + \underbrace{\dbinom{100}{100}}_{t_{100}} \\ \\ &= \underbrace{\left[ \dbinom{100}{0} x^{100} - \dbinom{100}{2} x^{98} + \cdots - \dbinom{100}{98} x^2 + \dbinom{100}{100} \right]}_{\displaystyle \sum_{r=0}^{50} {(-1)}^r t_{2r} } + i \underbrace{\left[ \dbinom{100}{1} x^{99} - \dbinom{100}{3} x^{97} + \cdots - \dbinom{100}{99} x \right]}_{\displaystyle \sum_{r=0}^{49} {(-1)}^r t_{2r+1} } \\ \\ &= \mathfrak{R} {(x+i)}^{100} + i \times \mathfrak{I} {(x+i)}^{100} \end{aligned}$

Thus the denominator is

$\begin{aligned} {\left[ \mathfrak{R} {(x+i)}^{100} \right]}^2 + {\left[ \mathfrak{I} {(x+i)}^{100} \right]}^2 &= {\left| {(x+i)}^{100} \right|}^2 \\ &= {\left({\left| x+i \right|}^2 \right)}^{100} \\ &= {\left({\left| x+a \right|}^2 \right)}^{100} \\ &= {(x^2 + a^2)}^{100} \end{aligned}$

Therefore, the integration is

$\displaystyle \int \dfrac{2x}{{(a^2+x^2)}^{100}} \,dx$

Which comes out to be

$C + \dfrac{-1}{99 {(x^2+a^2)}^{99}}$

which satisfies the rest of the conditions as stated in the problem.

Thus

$\alpha + \beta + \eta + \gamma = (-1) + 99 + 2 + 99 = \boxed{199}$