intlimits_0^{frac{1}{2}} frac{1}{1 - x^2} ln | vedclass

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(A) माना $I = \int\limits_0^{\frac{1}{2}} \frac{1}{1 - x^2} \ln \left( \frac{1 + x}{1 - x} \right) dx$. प्रतिस्थापन $t = \ln \left( \frac{1 + x}{1 - x

Vedclass · Accepted Answer

$\frac{1}{4} \ln^2 \left( \frac{1}{3} \right)$

Vedclass · Answer

(A) माना $I = \int\limits_0^{\frac{1}{2}} \frac{1}{1 - x^2} \ln \left( \frac{1 + x}{1 - x} \right) dx$. प्रतिस्थापन $t = \ln \left( \frac{1 + x}{1 - x} \right) = \ln(1 + x) - \ln(1 - x)$ का उपयोग करें। तब $dt = \left( \frac{1}{1 + x} - \frac{-1}{1 - x} \right) dx = \left( \frac{1 - x + 1 + x}{1 - x^2} \right) dx = \frac{2}{1 - x^2} dx$. अतः,$\frac{dx}{1 - x^2} = \frac{1}{2} dt$. जब $x = 0$,तब $t = \ln(1) = 0$. जब $x = \frac{1}{2}$,तब $t = \ln \left( \frac{1 + 1/2}{1 - 1/2} \right) = \ln \left( \frac{3/2}{1/2} \right) = \ln 3$. इसलिए,$I = \int\limits_0^{\ln 3} t \cdot \frac{1}{2} dt = \frac{1}{2} \left[ \frac{t^2}{2} \right]_0^{\ln 3} = \frac{1}{4} \ln^2 3$. चूंकि $\ln^2(1/3) = (\ln 1 - \ln 3)^2 = (-\ln 3)^2 = \ln^2 3$,अतः उत्तर $\frac{1}{4} \ln^2 3$ या $\frac{1}{4} \ln^2 \left( \frac{1}{3} \right)$ है।

$\int\limits_0^{\frac{1}{2}} \frac{1}{1 - x^2} \ln \left( \frac{1 + x}{1 - x} \right) dx$ का मान ज्ञात कीजिए।

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