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The value of $\int \frac{\left(x^{2}+1\right)}{x^{4}+x^{2}+1} d x$ is
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$\frac{1}{\sqrt{3}} \tan ^{-1}\left\{\frac{x-1 / x}{\sqrt{3}}\right\}+C$
Let $I=\int \frac{x^{2}+1}{x^{4}+x^{2}+1} d x$
$\begin{aligned}=& \int \frac{\left(1+\frac{1}{x^{2}}\right)}{x^{2}+1+\frac{1}{x^{2}}} d x \\=& \int \frac{\left(1+\frac{1}{x^{2}}\right)}{\left(x-\frac{1}{x}\right)^{2}+(\sqrt{3})^{2}} d x \\=& \int \frac{d t}{(\sqrt{3})^{2}+t^{2}} \\ \quad\left[\text { let } t=x-\frac{1}{x} \Rightarrow d t=\left(1+\frac{1}{x^{2}}\right) d x\right] \\=& \frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{t}{\sqrt{3}}\right)+C \\=& \frac{1}{\sqrt{3}} \tan ^{-1}\left\{\frac{1}{\sqrt{3}}\left(x-\frac{1}{x}\right)\right\}+C \end{aligned}$
$\begin{aligned}=& \int \frac{\left(1+\frac{1}{x^{2}}\right)}{x^{2}+1+\frac{1}{x^{2}}} d x \\=& \int \frac{\left(1+\frac{1}{x^{2}}\right)}{\left(x-\frac{1}{x}\right)^{2}+(\sqrt{3})^{2}} d x \\=& \int \frac{d t}{(\sqrt{3})^{2}+t^{2}} \\ \quad\left[\text { let } t=x-\frac{1}{x} \Rightarrow d t=\left(1+\frac{1}{x^{2}}\right) d x\right] \\=& \frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{t}{\sqrt{3}}\right)+C \\=& \frac{1}{\sqrt{3}} \tan ^{-1}\left\{\frac{1}{\sqrt{3}}\left(x-\frac{1}{x}\right)\right\}+C \end{aligned}$
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