Question

Evaluate \( \int \frac{x^2}{x^4+1} dx \).

Hint:

Remember the identity \( x^2 + \frac{1}{x^2} = (x \pm \frac{1}{x})^2 \mp 2 \). Choosing between \( + \) or \( - \) depends on whether the numerator is \( 1 \mp \frac{1}{x^2} \).

Answer 1

Step 1: Understanding the Concept:
This type of integral is solved by rewriting the numerator to split the integral into two standard forms.
Step 2: Detailed Explanation:
Write \( x^2 = \frac{1}{2} [ (x^2 + 1) + (x^2 - 1) ] \).
\[ I = \frac{1}{2} \int \frac{x^2 + 1}{x^4 + 1} dx + \frac{1}{2} \int \frac{x^2 - 1}{x^4 + 1} dx \] Divide both numerator and denominator by \( x^2 \):
\[ I = \frac{1}{2} \int \frac{1 + 1/x^2}{x^2 + 1/x^2} dx + \frac{1}{2} \int \frac{1 - 1/x^2}{x^2 + 1/x^2} dx \] For Part 1: Let \( t = x - 1/x \implies dt = (1 + 1/x^2)dx \). Denominator is \( t^2 + 2 \).
For Part 2: Let \( u = x + 1/x \implies du = (1 - 1/x^2)dx \). Denominator is \( u^2 - 2 \).
\[ I = \frac{1}{2} \int \frac{dt}{t^2 + 2} + \frac{1}{2} \int \frac{du}{u^2 - 2} \] \[ I = \frac{1}{2\sqrt{2}} \tan^{-1}\left(\frac{t}{\sqrt{2}}\right) + \frac{1}{4\sqrt{2}} \log \left| \frac{u-\sqrt{2}}{u+\sqrt{2}} \right| \] Substituting \( t \) and \( u \):
\[ I = \frac{1}{2\sqrt{2}} \tan^{-1}\left(\frac{x^2-1}{x\sqrt{2}}\right) + \frac{1}{4\sqrt{2}} \log \left| \frac{x^2 - \sqrt{2}x + 1}{x^2 + \sqrt{2}x + 1} \right| + C \].
Step 3: Final Answer:
The evaluated integral is \( \frac{1}{2\sqrt{2}} \tan^{-1}\left(\frac{x^2-1}{x\sqrt{2}}\right) + \frac{1}{4\sqrt{2}} \log \left| \frac{x^2 - \sqrt{2}x + 1}{x^2 + \sqrt{2}x + 1} \right| + C \).