intdx/((x^(1/2)-x^(1/4)) =কত?

ধরি, \(x = u^4\)

সুতরাং, \(dx = 4u^3 du\)

অতএব, \(\int \frac{dx}{x^{1/2} - x^{1/4}} = \int \frac{4u^3 du}{u^2 - u}\)

= \(4 \int \frac{u^3}{u(u-1)} du\)

= \(4 \int \frac{u^2}{u-1} du\)

এখন, \(u^2 = (u-1)(u+1) + 1\)

সুতরাং, \(4 \int \frac{u^2}{u-1} du = 4 \int \frac{(u-1)(u+1) + 1}{u-1} du\)

= \(4 \int (u+1 + \frac{1}{u-1}) du\)

= \(4 [\frac{u^2}{2} + u + ln|u-1|] + c\)

= \(2u^2 + 4u + 4ln|u-1| + c\)

যেহেতু, \(x = u^4\), সুতরাং \(u = x^{1/4}\)

অতএব, নির্ণেয় সমাধান:

\(2(x^{1/4})^2 + 4x^{1/4} + 4ln|x^{1/4} - 1| + c\)

= \(2x^{1/2} + 4x^{1/4} + 4ln|x^{1/4} - 1| + c\)

= \(2\sqrt{x} + 4\sqrt[4]{x} + ln(x^{1/4} - 1)^4 + c\)

সুতরাং, \(\int \frac{dx}{x^{1/2} - x^{1/4}} = 2\sqrt{x} + 4\sqrt[4]{x} + ln(\sqrt[4]{x} - 1)^4 + c\) 😃