$\int \frac{ax^2 + 2bx + c}{(Ax^2 + 2Bx + C)^2} dx $ (where $B^2\ne AC$) is a rational function then which of the condition must satisfy

Question

The options given are

(a) $2Bb = Ac + aC$

(b) $Aa + Bb = Cc$

(c) $Bb = aC + cA$

(d) $abA + cbC = 2Bac$

My attempt :

I assumed the above integral as $\frac{p(x)}{q(x)} $and differentiated both sides with respect to $x$.

So we get

$\frac{ax^2 + 2bx + c}{(Ax^2 + 2Bx + C)^2} $ =$ \frac{q(x).p'(x) - q'(x).p(x)}{q(x)^2}$ (quotient rule)

By comparison $q(x) = Ax^2 + 2Bx + C$

And assuming $p(x) = lx^2 + mx + n$

I get an expression for $l$ ,$m$ and $n$. (I don't see how it helps)

Is this the way to go? Or is there a better method?

Answer 1

The polynomial $p(x)$ should be of degree (at most) one: $p(x)=mx+n$. From $$ m(Ax^2+2Bx+C)-(mx+n)(2Ax+2B)=ax^2+2bx+c $$ you get a linear system in two unknowns, with three equations.

Thus you will get a relation between $a,b,c$ and $A,B,C$. Solve two equations for $m$ and $n$ and substitute in the third one.

If you try with a (formally) degree $2$ polynomial $p(x)=lx^2+mx+n$, you get $$ p'(x)q(x)-p(x)q'(x)=(2Bl- Am)x^2 + (2Cl - 2An)x + (Cm - 2Bn) $$

? (2*l*x+m)*(A*x^2+2*B*x+C)-(l*x^2+m*x+n)*(2*A*x+2*B)
%1 = (2*B*l - A*m)*x^2 + (2*C*l - 2*n*A)*x + (C*m - 2*n*B)

This leads to the linear system \begin{cases} 2Bl-Am=a \\[4px] 2Cl-2An=2b \\[4px] Cm-2Bn=c \end{cases} Multiply the first equation by $C$, the second equation by $B$ and subtract to get $$ ACm-2ABn=2bB-aC $$ From the third equation you get then $Ac=2bB-aC$. Solving for $l$ yields $l=0$.