Question

\[\int\frac{x}{4 + x^4} \text{ dx }\] is equal to

Options

• \[\frac{1}{4} \tan^{- 1} x^2 + C\]

• \[\frac{1}{4} \tan^{- 1} \left( \frac{x^2}{2} \right)\]

• \[\frac{1}{2} \tan^{- 1} \left( \frac{x^2}{2} \right)\]

• none of these

Answer 1

 \[\frac{1}{4} \tan^{- 1} \left( \frac{x^2}{2} \right)\]

\[\text{ Let  I } = \int\frac{x}{4 + x^4}dx\]

\[ = \int\frac{x \text{ dx}}{2^2 + \left( x^2 \right)^2}\]

\[\text{ Putting  x}^2 = t\]

\[ \Rightarrow 2x \text{ dx} = dt\]

\[ \Rightarrow x \text{ dx } = \frac{dt}{2}\]

\[ \therefore I = \frac{1}{2}\int\frac{dt}{2^2 + t^2}\]

\[ = \frac{1}{2} \times \frac{1}{2} \tan^{- 1} \left( \frac{t}{2} \right) + C \left( \because \int\frac{1}{a^2 + x^2} = \frac{1}{a} \tan^{- 1} \frac{x}{a} \right)\]

\[ = \frac{1}{4} \tan^{- 1} \left( \frac{x^2}{2} \right) + C \left( \because t = x^2 \right)\]