int1/x^2sin(1/x)dx এর মান কত?

Explanation:

Another Explanation (5): সমাধান: ধরি, \(u = \frac{1}{x}\) তাহলে, \(\frac{du}{dx} = -\frac{1}{x^2}\) সুতরাং, \(du = -\frac{1}{x^2} dx\) অতএব, \(\frac{1}{x^2} dx = -du\) এখন, \(\int \frac{1}{x^2} \sin\left(\frac{1}{x}\right) dx = \int \sin(u) (-du)\) \( = -\int \sin(u) du\) \( = - (-\cos(u)) + c\) \( = \cos(u) + c\) \( = \cos\left(\frac{1}{x}\right) + c\) সুতরাং, \(\int \frac{1}{x^2} \sin\left(\frac{1}{x}\right) dx = \cos\left(\frac{1}{x}\right) + c\) উত্তর: \(\cos\left(\frac{1}{x}\right) + c\) 🎉