Consider $$\phi(a,b,t) = a^4 - 5a^2 +b^2 + 5t^2 -4bt -2t + \frac{33}{4}$$ where $a,b,t ∈ R$.Given that $f(t)$ and $g(b)$ are the minimum values of $\phi(a,b,t)$.
Based on this statement there are two questions:
$\int \frac{dx}{f(x)} $ is $$ (a) \tan^{-1}(x-1)+c \\ (b) \frac{1}{2}tan^{-1}\left(\frac{x-1}{2}\right) + c \\ (c) ln \left[x-1+\sqrt{x^2-2x+2}\right]+c \\ (d) \frac{1}{2}ln\left|\frac{x-1}{x+1}\right| + c $$
If $\int{e^xg(x)}dx = e^x(Ax^2 + Bx + C) + D$, where $D$ is constant of integration then $(A+B+C)$ is equal to $$ (a)2 \\ (b) \frac{14}{5} \\ (c) 5 \\ (d) \frac{19}{5} $$
My attempt : Since the function $\phi$ is a function of three variables, I didn't know how to proceed with the question. Any hint will be very helpful.
To help you with the first, consider $$\phi(a,b,t) = a^4 - 5a^2 +b^2 + 5t^2 -4bt -2t + \frac{33}{4}$$ from which $$\frac{\partial}{\partial a}\phi(a,b,t) =4 a^3-10 a$$ $$\frac{\partial}{\partial b}\phi(a,b,t) =2 b-4 t$$ Since you look for an extremum, set these partial derivatives equal to $0$ and solve these two equations for $(a,b)$. Because of the first, you will find three solutions for $a$ and, for each of them, $b=2t$. You need to find which solution of $a$ (let us call it $a_*$ gives the minimum value of $\phi(a,b,t)$.
So $$f(t)=t^2-2 t+a_*^4-5 a_*^2+\frac{33}{4}$$ Complete the square and go on.
I am sure that you can take it from here.