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Find A(x) in ∫((x²−x+1)/(x²+1))e^(cot⁻¹x)dx | JEE

JEE Maths question with a full step-by-step solution.

Question
If ∫x2−x+1x2+1 ecot⁡−1x dx=A(x) ecot⁡−1x+C\displaystyle\int \dfrac{x^{2}-x+1}{x^{2}+1}\,e^{\cot^{-1}x}\,dx=A(x)\,e^{\cot^{-1}x}+C, then A(x)A(x) equals
A−x-x
B1−x\sqrt{1-x}
Cxxcorrect
D1+x\sqrt{1+x}
Solution
Step 1: Differentiate both sides. The left side gives x2−x+1x2+1 ecot⁡−1x\dfrac{x^{2}-x+1}{x^{2}+1}\,e^{\cot^{-1}x}, while the right side, using the product rule, gives
ddx(A(x) ecot⁡−1x)=A′(x) ecot⁡−1x+A(x)⋅ddxecot⁡−1x\dfrac{d}{dx}\big(A(x)\,e^{\cot^{-1}x}\big)=A'(x)\,e^{\cot^{-1}x}+A(x)\cdot\dfrac{d}{dx}e^{\cot^{-1}x}
 Step 2: Compute ddxecot⁡−1x=ecot⁡−1x⋅(−11+x2)\dfrac{d}{dx}e^{\cot^{-1}x}=e^{\cot^{-1}x}\cdot\left(-\dfrac{1}{1+x^{2}}\right), so the right side is
ecot⁡−1x[A′(x)−A(x)1+x2]e^{\cot^{-1}x}\left[A'(x)-\dfrac{A(x)}{1+x^{2}}\right]
 Step 3: Cancel ecot⁡−1xe^{\cot^{-1}x} and multiply through by (x2+1)(x^{2}+1):
x2−x+1=(x2+1)A′(x)−A(x)x^{2}-x+1=(x^{2}+1)A'(x)-A(x)
 Step 4: The right side must be a quadratic, so try the linear A(x)=xA(x)=x (hence A′(x)=1A'(x)=1):
(x2+1)(1)−x=x2+1−x=x2−x+1✓(x^{2}+1)(1)-x=x^{2}+1-x=x^{2}-x+1\quad\checkmark
∴A(x)=x\therefore A(x)=x
 Correct answer: (3)
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