int_0^1(3(tan^-1x)^2)/(1+x^2)dx- এর মান কত?

ধরি, \(I = \int_{0}^{1} \frac{3(\tan^{-1}x)^2}{1+x^2} dx\)

ধরি, \(u = \tan^{-1}x\)

তাহলে, \(\frac{du}{dx} = \frac{1}{1+x^2}\)

সুতরাং, \(du = \frac{dx}{1+x^2}\)

যখন \(x = 0\), তখন \(u = \tan^{-1}(0) = 0\)

যখন \(x = 1\), তখন \(u = \tan^{-1}(1) = \frac{\pi}{4}\)

অতএব, \(I = \int_{0}^{\frac{\pi}{4}} 3u^2 du\)

\(I = 3 \int_{0}^{\frac{\pi}{4}} u^2 du\)

\(I = 3 \left[ \frac{u^3}{3} \right]_{0}^{\frac{\pi}{4}}\)

\(I = 3 \left[ \frac{(\frac{\pi}{4})^3}{3} - \frac{0^3}{3} \right]\)

\(I = 3 \left[ \frac{\pi^3}{3 \cdot 4^3} \right]\)

\(I = \frac{\pi^3}{64}\)

সুতরাং, \(\int_{0}^{1} \frac{3(\tan^{-1}x)^2}{1+x^2} dx = \frac{\pi^3}{64}\) 🥳