JEE Main 2013 (Online) 22nd AprilQuestion 1
If $\int\frac{x^{2}-x+1}{x^{2}+1}e^{\cot^{-1}x}dx=A(x)e^{\cot^{-1}x}+C$, then $A(x)$ is equal to :
Indefinite Integration
Solution
Concept. Try $A(x)=x$: the product rule should give back the integrand. Formulas used. (1) $\frac{d}{dx}\cot^{-1}x=-\frac{1}{1+x^2}$Step 1 - Differentiate. By the product rule and (1), $$\frac{d}{dx}\Big(xe^{\cot^{-1}x}\Big)=e^{\cot^{-1}x}\Big(1-\frac{x}{1+x^2}\Big).$$ Step 2 - Simplify. $1-\frac{x}{1+x^2}=\frac{x^2-x+1}{x^2+1}$, which is the integrand's factor. So $A(x)=x$. Answer: (B)