int(x^2-1)/(x^2-4) dx এর মান কোনটি?

Explanation:

Another Explanation (5): সমাধান: \[ \int \frac{x^2 - 1}{x^2 - 4} dx \] প্রথমে, অপ্রকৃত ভগ্নাংশটিকে প্রকৃত ভগ্নাংশে পরিণত করি: \[ \frac{x^2 - 1}{x^2 - 4} = \frac{x^2 - 4 + 3}{x^2 - 4} = 1 + \frac{3}{x^2 - 4} \] তাহলে, \[ \int \frac{x^2 - 1}{x^2 - 4} dx = \int \left(1 + \frac{3}{x^2 - 4}\right) dx = \int 1 dx + \int \frac{3}{x^2 - 4} dx \] এখন, \(\int 1 dx = x + C_1\)। দ্বিতীয় ইন্টিগ্রালটি হলো: \[ \int \frac{3}{x^2 - 4} dx = 3 \int \frac{1}{x^2 - 4} dx = 3 \int \frac{1}{(x - 2)(x + 2)} dx \] আংশিক ভগ্নাংশে প্রকাশ করে পাই: \[ \frac{1}{(x - 2)(x + 2)} = \frac{A}{x - 2} + \frac{B}{x + 2} \] \[ 1 = A(x + 2) + B(x - 2) \] যদি \(x = 2\) হয়, তবে \(1 = 4A \implies A = \frac{1}{4}\)। যদি \(x = -2\) হয়, তবে \(1 = -4B \implies B = -\frac{1}{4}\)। সুতরাং, \[ \frac{1}{(x - 2)(x + 2)} = \frac{1/4}{x - 2} - \frac{1/4}{x + 2} = \frac{1}{4} \left(\frac{1}{x - 2} - \frac{1}{x + 2}\right) \] তাহলে, \[ 3 \int \frac{1}{x^2 - 4} dx = 3 \int \frac{1}{4} \left(\frac{1}{x - 2} - \frac{1}{x + 2}\right) dx = \frac{3}{4} \int \left(\frac{1}{x - 2} - \frac{1}{x + 2}\right) dx \] \[ = \frac{3}{4} \left(\int \frac{1}{x - 2} dx - \int \frac{1}{x + 2} dx\right) = \frac{3}{4} \left(\ln|x - 2| - \ln|x + 2|\right) + C_2 \] \[ = \frac{3}{4} \ln\left|\frac{x - 2}{x + 2}\right| + C_2 \] অতএব, \[ \int \frac{x^2 - 1}{x^2 - 4} dx = x + \frac{3}{4} \ln\left|\frac{x - 2}{x + 2}\right| + C \] যেখানে \(C = C_1 + C_2\)। সুতরাং, উত্তর: ✅