Question

If $\int \left(\frac{x^{2} - x + 1}{x^{2} + 1}\right)\cdot e^{\left(cot\right)^{- 1} x}dx$ $=A\left(x\right)\cdot e^{\left(cot\right)^{- 1} x}+C$ , where $C$ is constant of integration, then $A\left(x\right)$ is equal to

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

Answer 1

Answer: (C)

Method-I:
Differentiate both sides with respect to $x$ , we get
$\frac{\left(x^{2} + 1\right) - x}{x^{2} + 1}=\frac{\left(x^{2} + 1\right) A^{'} \left(\right. x \left.\right) - A \left(\right. x \left.\right)}{x^{2} + 1}$
$\therefore A\left(\right.x\left.\right)=x$ Ans.
Method-II:
Let $I=\int \left(\frac{x^{2} - x + 1}{1 + x^{2}}\right)\cdot e^{\left(cot\right)^{- 1} x}dx$
Put $cot^{- 1}x=t$ , we get
$I=\int e^{t}\left(cot t - \left(cosec\right)^{2} t\right)dt=e^{t}\cdot cot\left(\right.t\left.\right)+c$
$=x\cdot e^{cot^{- 1} x}+c$