Advertisements
Advertisements
प्रश्न
\[\int\frac{x}{4 + x^4} \text{ dx }\] is equal to
पर्याय
\[\frac{1}{4} \tan^{- 1} x^2 + C\]
\[\frac{1}{4} \tan^{- 1} \left( \frac{x^2}{2} \right)\]
\[\frac{1}{2} \tan^{- 1} \left( \frac{x^2}{2} \right)\]
none of these
MCQ
Advertisements
उत्तर
\[\frac{1}{4} \tan^{- 1} \left( \frac{x^2}{2} \right)\]
\[\text{ Let I } = \int\frac{x}{4 + x^4}dx\]
\[ = \int\frac{x \text{ dx}}{2^2 + \left( x^2 \right)^2}\]
\[\text{ Putting x}^2 = t\]
\[ \Rightarrow 2x \text{ dx} = dt\]
\[ \Rightarrow x \text{ dx } = \frac{dt}{2}\]
\[ \therefore I = \frac{1}{2}\int\frac{dt}{2^2 + t^2}\]
\[ = \frac{1}{2} \times \frac{1}{2} \tan^{- 1} \left( \frac{t}{2} \right) + C \left( \because \int\frac{1}{a^2 + x^2} = \frac{1}{a} \tan^{- 1} \frac{x}{a} \right)\]
\[ = \frac{1}{4} \tan^{- 1} \left( \frac{x^2}{2} \right) + C \left( \because t = x^2 \right)\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
