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 :
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 :
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Differentiate \(x\,e^{\cot^{-1}x}\) and compare with the integrand, or substitute \(x=\cot t\) and use \(\int e^t(f+f')\,dt\).
Updated On: Oct 31, 2026
- \(-x\)
- \(x\)
- \(\sqrt{1-x}\)
- \(\sqrt{1+x}\)
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The Correct Option is B
Solution and Explanation
Try \(A(x)=x\). Using \(\frac{d}{dx}\cot^{-1}x=-\frac{1}{1+x^2}\):
\(\frac{d}{dx}\left[x\,e^{\cot^{-1}x}\right]=e^{\cot^{-1}x}+x\,e^{\cot^{-1}x}\cdot\left(-\frac{1}{1+x^2}\right)=e^{\cot^{-1}x}\cdot\frac{x^2+1-x}{x^2+1}\).
This equals the integrand \(\frac{x^2-x+1}{x^2+1}e^{\cot^{-1}x}\), so the integral is \(x\,e^{\cot^{-1}x}+C\) and \(A(x)=x\).
Answer: (2) \(x\)
\(\frac{d}{dx}\left[x\,e^{\cot^{-1}x}\right]=e^{\cot^{-1}x}+x\,e^{\cot^{-1}x}\cdot\left(-\frac{1}{1+x^2}\right)=e^{\cot^{-1}x}\cdot\frac{x^2+1-x}{x^2+1}\).
This equals the integrand \(\frac{x^2-x+1}{x^2+1}e^{\cot^{-1}x}\), so the integral is \(x\,e^{\cot^{-1}x}+C\) and \(A(x)=x\).
Answer: (2) \(x\)
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