समाकल int limits_{0}^{1} frac{sqrt{x} , dx}{( | vedclass

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Question

(A) $I = \int_{0}^{1} \frac{\sqrt{x}}{(1+x)(1+3x)(3+x)} \, dx$माना $x = t^2$,तब $dx = 2t \, dt$। जब $x=0, t=0$ और जब $x=1, t=1$।$I = \int_{0}^{1} \fra

Vedclass · Accepted Answer

$\frac{\pi}{8}\left(1-\frac{\sqrt{3}}{2}\right)$

Vedclass · Answer

(A) $I = \int_{0}^{1} \frac{\sqrt{x}}{(1+x)(1+3x)(3+x)} \, dx$माना $x = t^2$,तब $dx = 2t \, dt$। जब $x=0, t=0$ और जब $x=1, t=1$।$I = \int_{0}^{1} \frac{t(2t)}{(t^2+1)(1+3t^2)(3+t^2)} \, dt = \int_{0}^{1} \frac{2t^2}{(t^2+1)(3t^2+1)(t^2+3)} \, dt$आंशिक भिन्नों का उपयोग करके,हम समाकल्य को इस प्रकार लिख सकते हैं:$\frac{2t^2}{(t^2+1)(3t^2+1)(t^2+3)} = \frac{A}{t^2+1} + \frac{B}{3t^2+1} + \frac{C}{t^2+3}$गुणांकों को हल करने पर,हमें $A = \frac{1}{2}, B = -\frac{3}{8}, C = -\frac{1}{8}$ प्राप्त होता है।$I = \frac{1}{2} \int_{0}^{1} \frac{dt}{t^2+1} - \frac{3}{8} \int_{0}^{1} \frac{dt}{3t^2+1} - \frac{1}{8} \int_{0}^{1} \frac{dt}{t^2+3}$$I = \frac{1}{2} [	an^{-1}(t)]_{0}^{1} - \frac{3}{8} \cdot \frac{1}{\sqrt{3}} [	an^{-1}(\sqrt{3}t)]_{0}^{1} - \frac{1}{8} \cdot \frac{1}{\sqrt{3}} [	an^{-1}(\frac{t}{\sqrt{3}})]_{0}^{1}$$I = \frac{1}{2} (\frac{\pi}{4}) - \frac{\sqrt{3}}{8} (\frac{\pi}{3}) - \frac{\sqrt{3}}{24} (\frac{\pi}{6})$गणना करने पर,अंतिम उत्तर $I = \frac{\pi}{8}(1 - \frac{\sqrt{3}}{2})$ प्राप्त होता है।

समाकल $\int \limits_{0}^{1} \frac{\sqrt{x} \, dx}{(1+x)(1+3 x)(3+x)}$ का मान है:

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