intdx/(xsqrt(x^4-1)) এর মান কত হবে?

ধরি, \(I = \int \frac{dx}{x\sqrt{x^4-1}}\)

এখানে, \(x^2 = t\) ধরলে, \(2xdx = dt \Rightarrow dx = \frac{dt}{2x}\)

সুতরাং, \(I = \int \frac{dt}{2x^2\sqrt{x^4-1}} = \int \frac{dt}{2t\sqrt{t^2-1}}\)

এখন, \(t = \sec\theta\) ধরলে, \(dt = \sec\theta \tan\theta d\theta\)

তাহলে, \(I = \int \frac{\sec\theta \tan\theta d\theta}{2\sec\theta \sqrt{\sec^2\theta - 1}} = \int \frac{\sec\theta \tan\theta d\theta}{2\sec\theta \tan\theta} = \int \frac{1}{2} d\theta\)

\(= \frac{1}{2} \theta + C = \frac{1}{2} \sec^{-1}t + C\)

\(= \frac{1}{2} \sec^{-1}(x^2) + C\)

\(= \frac{1}{2} \cos^{-1}\left(\frac{1}{x^2}\right) + C\) 🥳