Question

If $y^2=3x^2+2x+1$ & integration $I_n$ is defined as $I_n=\int \frac{x^n}{y}dx$ where $A{I}_{10}+B{I}_9+C{I}_8=x^{9}y$ then

Answer 1

$y^2=3x^2+2x+1$
$I_n=\int \frac{x^n}{y}dx$
$A{I}_{10}+B{I}_9+C{I}_8=x^{9}y$
$A\int \frac{x^{10}}{y}dx+B\int \frac{x^9}{y}dx+C\int \frac{x^8}{y}dx=x^9.y$
differtiate both sides
$\frac{A{x}^{10}}{y}+\frac{B{x}^9}{y}+\frac{C{x}^8}{y}=x^9\frac{dy}{dx}+y.9.x^8$
$A{x}^{10}+B{x}^9+C{x}^8=x^9.y\frac{dy}{dx}+9.y^2{x}^8$
$A{x}^{10}+B{x}^9+C{x}^8=x^9(3x+1)+9x^8(3x^2+2x+1)$
Compare the cofficients
$\Rightarrow A=30$
$B=19$
$C=9$
$\Rightarrow x-int$ of $Cx+By+A=0\Rightarrow \left|\frac{3A}{C}\right|=10$