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

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, \(f(x) = \sqrt{1+x^2}\) 🥳