Question

If \(\int \frac{x^2-x+1}{x^2+1}\,e^{\cot^{-1}x}\,dx = A(x)\,e^{\cot^{-1}x}+C\), then \(A(x)\) is equal to :

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

Hint:

Differentiate \(x\,e^{\cot^{-1}x}\) and compare with the integrand, or substitute \(x=\cot t\) and use \(\int e^t(f+f')\,dt\).

Answer 1

The Correct Option is B

Put \(x=\cot t\), so \(dx=-\csc^2t\,dt\) and \(x^2+1=\csc^2t\).
Integral \(=\int\frac{\cot^2t-\cot t+1}{\csc^2t}\,e^t\,(-\csc^2t)\,dt=-\int(\cot^2t+1-\cot t)\,e^t\,dt=-\int(\csc^2t-\cot t)\,e^t\,dt\).
\(=\int e^t(\cot t-\csc^2t)\,dt\). Since \(\frac{d}{dt}(\cot t)=-\csc^2t\), this is of the form \(\int e^t(f(t)+f'(t))\,dt=e^tf(t)\) with \(f=\cot t\).
So the integral \(=e^t\cot t+C=x\,e^{\cot^{-1}x}+C\), giving \(A(x)=x\).
Answer: (2) \(x\)