intdx/(sqrt(-2x^2+4x+1)) এর মান কোনটি?

সমাধান:

ধরি, \(I = \int \frac{dx}{\sqrt{-2x^2+4x+1}}\)

এখন, \(-2x^2+4x+1 = -2(x^2-2x) + 1\)

\(= -2(x^2-2x+1-1) + 1\)

\(= -2((x-1)^2 - 1) + 1\)

\(= -2(x-1)^2 + 2 + 1\)

\(= 3 - 2(x-1)^2\)

\(= 3\left[1 - \frac{2}{3}(x-1)^2\right]\)

সুতরাং, \(I = \int \frac{dx}{\sqrt{3\left[1 - \frac{2}{3}(x-1)^2\right]}}\)

\(= \frac{1}{\sqrt{3}} \int \frac{dx}{\sqrt{1 - \left(\sqrt{\frac{2}{3}}(x-1)\right)^2}}\)

ধরি, \(u = \sqrt{\frac{2}{3}}(x-1)\)

তাহলে, \(\frac{du}{dx} = \sqrt{\frac{2}{3}}\)

\(dx = \sqrt{\frac{3}{2}}du\)

সুতরাং, \(I = \frac{1}{\sqrt{3}} \int \frac{\sqrt{\frac{3}{2}}du}{\sqrt{1-u^2}}\)

\(= \frac{1}{\sqrt{3}} \cdot \sqrt{\frac{3}{2}} \int \frac{du}{\sqrt{1-u^2}}\)

\(= \frac{1}{\sqrt{2}} \int \frac{du}{\sqrt{1-u^2}}\)

\(= \frac{1}{\sqrt{2}} \sin^{-1}(u) + C\)

\(= \frac{1}{\sqrt{2}} \sin^{-1}\left(\sqrt{\frac{2}{3}}(x-1)\right) + C\)

অতএব, \(\int \frac{dx}{\sqrt{-2x^2+4x+1}} = \frac{1}{\sqrt{2}} \sin^{-1}\left(\sqrt{\frac{2}{3}}(x-1)\right) + C\) 🥳