Question
$\int \frac{x^{4}+1}{x^{6}+1} d x=\tan ^{-1} k_{1}-\frac{2}{3} \tan ^{-1} k_{2}+C$, where(A) $k_{1}=x+\frac{1}{x}$(B) $k_{2}=x^{3}$(C) $k_{1}=x-\frac{1}{v}$(D) $k_{2}=x^{4}$
Step 1
We can rewrite the numerator as $x^{4}+1 = x^{2}+1^{2} - 2x^{2}$ using the formula $a^{2}+b^{2} = (a+b)^{2} - 2ab$. Show more…
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