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$\int \frac{x^{2}+1}{x^{4}+x^{2}+1} d x=$
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Verified Answer
The correct answer is:
$\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{3}}\right)+c$
Let
$\begin{aligned} I &=\int \frac{x^{2}+1}{x^{4}+x^{2}+1} d x \\ &=\int \frac{x^{2}\left(1+\frac{1}{x^{2}}\right)}{x^{2}\left(x^{2}+1+\frac{1}{x^{2}}\right)} d x=\int \frac{1+\frac{1}{x^{2}}}{\left(x^{2}+\frac{1}{x^{2}}-2\right)+3} \\ I &=\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+(\sqrt{3})^{2}} d x \end{aligned}$
Put $x-\frac{1}{x}=t \Rightarrow\left(1+\frac{1}{x^{2}}\right) d x=d t$
$\therefore I=\int \frac{d t}{t^{2}+(\sqrt{3})^{2}}$
$=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{t}{\sqrt{3}}\right)+c=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{3}}\right)+c$
$\begin{aligned} I &=\int \frac{x^{2}+1}{x^{4}+x^{2}+1} d x \\ &=\int \frac{x^{2}\left(1+\frac{1}{x^{2}}\right)}{x^{2}\left(x^{2}+1+\frac{1}{x^{2}}\right)} d x=\int \frac{1+\frac{1}{x^{2}}}{\left(x^{2}+\frac{1}{x^{2}}-2\right)+3} \\ I &=\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+(\sqrt{3})^{2}} d x \end{aligned}$
Put $x-\frac{1}{x}=t \Rightarrow\left(1+\frac{1}{x^{2}}\right) d x=d t$
$\therefore I=\int \frac{d t}{t^{2}+(\sqrt{3})^{2}}$
$=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{t}{\sqrt{3}}\right)+c=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{3}}\right)+c$
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