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

Question

Aicharya · Accepted Answer

Let $\mathrm{I}=\int \frac{x^2-x+1}{x^2+1} \cdot e^{\cot ^{-1} x} d x$ Put $x=\cot t \Rightarrow-\operatorname{cosec}^2 t d t=d x$ Now, $1+\cot ^2 t=\operatorname{cosec}^2 t$ $	herefore$ $\mathrm{I}=\int \frac{e^t\left(\cot ^2 t-\cot t+1ight)}{\left(1+\cot ^2 tight)}\left(-\operatorname{cosec}^2 tight) d t$ $ \begin{aligned} & =-\int e^t\left(\operatorname{cosec}{ }^2 t-\cot tight) d t \ & =\int e^t\left(\cot t-\operatorname{cosec}^2 tight) d t \ & =e^t \cot t+\mathrm{C} \ & =e^{\cot ^{-1} x}(x)+\mathrm{C} \equiv \mathrm{A}(x) \cdot e^{\cot ^{-1} x}+\mathrm{C} \ \Rightarrow \mathrm{A}(x) & =x \end{aligned} $

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

Solution