Question

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

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

Answer 1

Answer: (A)

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