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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 :
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Verified Answer
The correct answer is:
$x$
$x$
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, $1+\cot ^2 t=\operatorname{cosec}^2 t$
$\therefore$
$\mathrm{I}=\int \frac{e^t\left(\cot ^2 t-\cot t+1\right)}{\left(1+\cot ^2 t\right)}\left(-\operatorname{cosec}^2 t\right) d t$
$$
\begin{aligned}
& =-\int e^t\left(\operatorname{cosec}{ }^2 t-\cot t\right) d t \\
& =\int e^t\left(\cot t-\operatorname{cosec}^2 t\right) d t \\
& =e^t \cot t+\mathrm{C} \\
& =e^{\cot ^{-1} x}(x)+\mathrm{C} \equiv \mathrm{A}(x) \cdot e^{\cot ^{-1} x}+\mathrm{C} \\
\Rightarrow \mathrm{A}(x) & =x
\end{aligned}
$$
Put $x=\cot t \Rightarrow-\operatorname{cosec}^2 t d t=d x$
Now, $1+\cot ^2 t=\operatorname{cosec}^2 t$
$\therefore$
$\mathrm{I}=\int \frac{e^t\left(\cot ^2 t-\cot t+1\right)}{\left(1+\cot ^2 t\right)}\left(-\operatorname{cosec}^2 t\right) d t$
$$
\begin{aligned}
& =-\int e^t\left(\operatorname{cosec}{ }^2 t-\cot t\right) d t \\
& =\int e^t\left(\cot t-\operatorname{cosec}^2 t\right) d t \\
& =e^t \cot t+\mathrm{C} \\
& =e^{\cot ^{-1} x}(x)+\mathrm{C} \equiv \mathrm{A}(x) \cdot e^{\cot ^{-1} x}+\mathrm{C} \\
\Rightarrow \mathrm{A}(x) & =x
\end{aligned}
$$
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