Explanation: 
Another Explanation (5):
সমাধান:
ধরি, \(I = \int \frac{x}{\sqrt{1+x^2}} dx\)
এখানে, \(1+x^2 = z\) ধরলে,
\(2x dx = dz\)
সুতরাং, \(x dx = \frac{1}{2} dz\)
তাহলে,
\(I = \int \frac{1}{\sqrt{z}} \cdot \frac{1}{2} dz\)
\(I = \frac{1}{2} \int z^{-\frac{1}{2}} dz\)
\(I = \frac{1}{2} \cdot \frac{z^{-\frac{1}{2}+1}}{-\frac{1}{2}+1} + c\)
\(I = \frac{1}{2} \cdot \frac{z^{\frac{1}{2}}}{\frac{1}{2}} + c\)
\(I = z^{\frac{1}{2}} + c\)
\(I = \sqrt{z} + c\)
z এর মান বসিয়ে পাই,
\(I = \sqrt{1+x^2} + c\)
সুতরাং, \(f(x) = \sqrt{1+x^2}\) 🥳
