Question

If \[ \int \frac{x^2-x+1}{x^2+1}\,e^{\cot^{-1}x}\,dx = A(x)e^{\cot^{-1}x}+C, \] then \(A(x)=\)

• A. \(-x\)
• B. \(x\)
• C. \(\sqrt{1-x}\)
• D. \(\sqrt{1+x}\)

Hint:

If the integral is given in the form \[ \int f(x)e^{g(x)}dx=A(x)e^{g(x)}+C, \] differentiate \(A(x)e^{g(x)}\) and compare with the integrand. In MCQs, checking the options is usually the fastest method.

Answer 1

The Correct Option is B

Concept: Use the method of inspection. Let \[ I=\int \frac{x^2-x+1}{x^2+1}\,e^{\cot^{-1}x}\,dx =A(x)e^{\cot^{-1}x}+C. \] Differentiate both sides and determine \(A(x)\).

Step 1: Differentiate the right-hand side. Since \[ \frac{d}{dx}\big(\cot^{-1}x\big) = -\frac{1}{1+x^2}, \] we get \[ \frac{d}{dx} \left( A(x)e^{\cot^{-1}x} \right) = e^{\cot^{-1}x} \left( A'(x)-\frac{A(x)}{1+x^2} \right). \] This must equal the integrand: \[ e^{\cot^{-1}x} \frac{x^2-x+1}{x^2+1}. \] Hence, \[ A'(x)-\frac{A(x)}{1+x^2} = \frac{x^2-x+1}{x^2+1}. \]

Step 2: Test the given options. Take \[ A(x)=x. \] Then \[ A'(x)=1. \] Therefore, \[ A'(x)-\frac{A(x)}{1+x^2} = 1-\frac{x}{1+x^2}. \] \[ = \frac{1+x^2-x}{1+x^2}. \] \[ = \frac{x^2-x+1}{x^2+1}. \] This exactly matches the integrand. Hence, \[ A(x)=x. \] Therefore, \[ \boxed{A(x)=x} \] \[ \boxed{\text{Answer = (B)}} \]