Question

If $I=\int \frac{a x^2+2 b x+c}{\left(A x^2+2 B x+C\right)^2} d x$ (where $B^2 \neq A C$ ) is a rational function then which one of the following condition must be necessary?

• A. $2 Bb = Ac + aC$
• B. $Aa + Bb = Cc$
• C. $Bb = aC + cA$
• D. $\frac{ A }{ c }+\frac{ C }{ a }=\frac{2 B }{ b }$

Answer 1

Answer: (A)

Let $\int \frac{a x^2+2 b x+c}{\left(A x^2+2 B x+C\right)^2} d x=\frac{f(x)}{g(x)}$ where $f(x)$ and $g(x)$ are polynomials ;
$\left( B ^2 \neq AC \Rightarrow Ax ^2+2 Bx + C\right.$ is not a perfect square $)$
differentiate w.r.t. $x$
$\frac{a x^2+2 b x+c}{\left(A x^2+2 B x+C\right)^2}=\frac{g(x) f^{\prime}(x)-f(x) g^{\prime}(x)}{g^2(x)}$
Hence $g(x) \cong A x^2+2 B x+C$
If $N ^{ r }$ of RHS in equation (1), has to be a quadratic function then $f ( x )$ must be linear function (think !)
$\text { i.e. } \quad f(x)=p x+q $
$\therefore p \left(A x ^2+2 Bx + C \right)-( px + q )(2 Ax +2 B ) \equiv ax ^2+2 bx + c$
Comparing coefficient of $x ^2$
$\text { Ap-2Ap }=a \Rightarrow-A p=a \Rightarrow p=\frac{-a}{A}$
$\text { coefficient of } x $
$2 B p-(2 B p+2 A q)=2 b \Rightarrow-A q=b \Rightarrow q=\frac{-b}{A} $
$\text { constant term }\quad pC -2 Bq = c$
$\text { substituting the value of } p \text { and } q $
$ \frac{-a C}{A}+\frac{2 B b}{A}=c \Rightarrow 2 B b-a C=A c \Rightarrow 2 B b=A c+a C$
$\text { Hence } 2 Bb =A c+a C $