int(x)/(sqrt(1+x^2))dx=f(x) হলে, f(x)=কত?

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

এখন, \(1+x^2 = z\) ধরলে, \(2x dx = dz\) হয়।

সুতরাং, \(x dx = \frac{1}{2} dz\).

তাহলে, \(I = \int \frac{1}{\sqrt{z}} \cdot \frac{1}{2} dz = \frac{1}{2} \int z^{-\frac{1}{2}} dz\).

\(\implies I = \frac{1}{2} \cdot \frac{z^{-\frac{1}{2}+1}}{-\frac{1}{2}+1} + c = \frac{1}{2} \cdot \frac{z^{\frac{1}{2}}}{\frac{1}{2}} + c\).

\(\implies I = \sqrt{z} + c\).

z এর মান বসিয়ে পাই, \(I = \sqrt{1+x^2} + c\).

অতএব, \(f(x) = \sqrt{1+x^2} + c\) 🎉।