Question

If $y^2=x^2-x+1$ and $I_n=\int \frac{x^n}{y}dx$ and $A{I}_3+B{I}_2+C{I}_1=x^{2}y$ then ordered triplet $A,B,C$ is

• A. $(\frac{1}{2},-\frac{1}{2},1)$
• B. $(3,1,0)$
• C. $(1,-1,2)$
• D. $(3,-\frac{5}{2},2)$

Answer 1

The correct option is A $(\frac{1}{2},-\frac{1}{2},1)$

We have given

$y^2=x^2-x+1$

$I_n=\int \frac{x^n}{y}dx$

And,

$A{I}_3+B{I}_2+C{I}_1=x^{2}y$

Order triplet $A,B,C$ $=\left(\frac{1}{2},\frac{-1}{2},1\right)$

Hence, the option $\left(A\right)$ is correct.