intx/(sqrt(1-x^4))dx= কত?

1/x cos^-1x^2+c

1/4 sin^-1x^2+c

-1/2 cos^-1x^2+c

1/2 sin^-1x^2+c

সঠিক উত্তরঃ D.

1/2 sin^-1x^2+c

Explanation:

Another Explanation (5): ```html

ধরি, \(x^2 = z\)

অতএব, \(2x dx = dz\)

সুতরাং, \(x dx = \frac{1}{2} dz\)

এখন, \(\int \frac{x}{\sqrt{1-x^4}} dx = \int \frac{1}{\sqrt{1-z^2}} \cdot \frac{1}{2} dz\)

\(= \frac{1}{2} \int \frac{1}{\sqrt{1-z^2}} dz\)

আমরা জানি, \(\int \frac{1}{\sqrt{1-x^2}} dx = \sin^{-1}x + c\)

সুতরাং, \(\frac{1}{2} \int \frac{1}{\sqrt{1-z^2}} dz = \frac{1}{2} \sin^{-1}z + c\)

z এর মান বসিয়ে পাই,

\(\frac{1}{2} \sin^{-1}(x^2) + c\)

অতএব, \(\int \frac{x}{\sqrt{1-x^4}} dx = \frac{1}{2} \sin^{-1}x^2 + c\) 🥳

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