यदि int frac{d x}{left(x^{2}+x+1right)^{2}}=a | vedclass

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(B) माना $I = \int \frac{dx}{(x^2+x+1)^2} = \int \frac{dx}{((x+1/2)^2 + 3/4)^2}$.$t = x + 1/2$ प्रतिस्थापित करने पर,$dt = dx$. अतः $I = \int \frac{dt}

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(B) माना $I = \int \frac{dx}{(x^2+x+1)^2} = \int \frac{dx}{((x+1/2)^2 + 3/4)^2}$.$t = x + 1/2$ प्रतिस्थापित करने पर,$dt = dx$. अतः $I = \int \frac{dt}{(t^2 + 3/4)^2}$.$t = \frac{\sqrt{3}}{2} 	an 	heta$ रखने पर,$dt = \frac{\sqrt{3}}{2} \sec^2 	heta d	heta$.$I = \int \frac{\frac{\sqrt{3}}{2} \sec^2 	heta d	heta}{(\frac{3}{4} 	an^2 	heta + 3/4)^2} = \int \frac{\frac{\sqrt{3}}{2} \sec^2 	heta d	heta}{\frac{9}{16} \sec^4 	heta} = \frac{8\sqrt{3}}{9} \int \cos^2 	heta d	heta$.$\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$ का उपयोग करने पर,$I = \frac{4\sqrt{3}}{9} \int (1 + \cos 2	heta) d	heta = \frac{4\sqrt{3}}{9} (	heta + \frac{\sin 2	heta}{2}) + C$.चूँकि $	an 	heta = \frac{2t}{\sqrt{3}} = \frac{2x+1}{\sqrt{3}}$,इसलिए $	heta = 	an^{-1}(\frac{2x+1}{\sqrt{3}})$.साथ ही,$\frac{\sin 2	heta}{2} = \frac{	an 	heta}{1 + 	an^2 	heta} = \frac{\frac{2x+1}{\sqrt{3}}}{1 + \frac{(2x+1)^2}{3}} = \frac{\sqrt{3}(2x+1)}{4(x^2+x+1)}$.अतः,$I = \frac{4\sqrt{3}}{9} 	an^{-1}(\frac{2x+1}{\sqrt{3}}) + \frac{1}{3} \frac{2x+1}{x^2+x+1} + C$.दिए गए रूप से तुलना करने पर,$a = \frac{4\sqrt{3}}{9}$ और $b = \frac{1}{3}$.अतः $9(\sqrt{3}a + b) = 9(\sqrt{3} \cdot \frac{4\sqrt{3}}{9} + \frac{1}{3}) = 9(\frac{4}{3} + \frac{1}{3}) = 15$.

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