સંકલન int_{-1/2}^{1/2} left[ left( frac{x+1}{ | vedclass

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Question

(C) ધારો કે $I = \int_{-1/2}^{1/2} \left[ \left( \frac{x+1}{x-1} \right)^2 + \left( \frac{x-1}{x+1} \right)^2 - 2 \right]^{1/2} dx$. વર્ગમૂળની અંદરનું

Vedclass · Accepted Answer

$4\log \left( \frac{4}{3} \right)$

Vedclass · Answer

(C) ધારો કે $I = \int_{-1/2}^{1/2} \left[ \left( \frac{x+1}{x-1} \right)^2 + \left( \frac{x-1}{x+1} \right)^2 - 2 \right]^{1/2} dx$. વર્ગમૂળની અંદરનું પદ $a^2 + b^2 - 2 = (a-b)^2$ સ્વરૂપનું છે,જ્યાં $a = \frac{x+1}{x-1}$ અને $b = \frac{x-1}{x+1}$ છે. તેથી,$\left( \frac{x+1}{x-1} - \frac{x-1}{x+1} \right)^2 = \left( \frac{(x+1)^2 - (x-1)^2}{(x-1)(x+1)} \right)^2 = \left( \frac{4x}{x^2-1} \right)^2$. આમ,$I = \int_{-1/2}^{1/2} \left| \frac{4x}{x^2-1} \right| dx = \int_{-1/2}^{1/2} \left| \frac{4x}{1-x^2} \right| dx$. વિધેય યુગ્મ હોવાથી,$I = 2 \int_{0}^{1/2} \frac{4x}{1-x^2} dx = 8 \int_{0}^{1/2} \frac{x}{1-x^2} dx$. ધારો કે $u = 1-x^2$,તો $du = -2x dx$,એટલે કે $x dx = -\frac{1}{2} du$. જ્યારે $x=0, u=1$; જ્યારે $x=1/2, u=3/4$. $I = 8 \int_{1}^{3/4} \frac{-1/2}{u} du = 4 \int_{3/4}^{1} \frac{1}{u} du = 4 [\log |u|]_{3/4}^{1} = 4 (\log 1 - \log(3/4)) = -4 \log(3/4) = 4 \log(4/3)$.

સંકલન $\int_{-1/2}^{1/2} \left[ \left( \frac{x+1}{x-1} \right)^2 + \left( \frac{x-1}{x+1} \right)^2 - 2 \right]^{1/2} dx$ નું મૂલ્ય શોધો.

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