(x^2 + 1)/(x^4 + 1) dx = Option 1: (x^4 + 1) + C Option 2: ((x^2 - 1)/(x^2 + 1)) + C Option 3: 1/√2 ^(-1)((x^2 - 1)/(√2 x)) + C Option 4: 1/√2 ^(-1)((x^2 + 1)/(√2 x)) + C

Answer: 1/√2 ^(-1)((x^2 - 1)/(√2 x)) + C. Solution: Divide numerator and denominator by x^2: (1 + 1/x^2)/(x^2 + 1/x^2) dx. Note x^2 + 1/x^2 = (x - 1/x)^2 + 2. Let t = x - 1/x, then dt = (1 + 1/x^2) dx. Integral becomes dt/(t^2 + 2) = 1/√2 ^(-1)(t/√2) + C = 1/√2 ^(-1)((x^2-1)/(√2 x)) + C.

Calculus

Integrals

Hard

$\int \frac{x^2 + 1}{x^4 + 1} dx =$

$\log(x^4 + 1) + C$

$\log\left(\frac{x^2 - 1}{x^2 + 1}\right) + C$

$\frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{x^2 - 1}{\sqrt{2} \, x}\right) + C$

$\frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{x^2 + 1}{\sqrt{2} \, x}\right) + C$

✅ Correct Option: 3

Divide numerator and denominator by $x^2$ : $\int \frac{1 + 1/x^2}{x^2 + 1/x^2} dx$ . Note $x^2 + 1/x^2 = (x - 1/x)^2 + 2$ . Let $t = x - 1/x$ , then $dt = (1 + 1/x^2) dx$ . Integral becomes $\int \frac{dt}{t^2 + 2} = \frac{1}{\sqrt{2}} \tan^{-1}(t/\sqrt{2}) + C = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{x^2-1}{\sqrt{2}\, x}\right) + C$ .

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Solution

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$\int \frac{x^2 + 1}{x^4 + 1} dx =$