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

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int\limits_0^{\frac{1}{2}} \frac{1}{1 - x^2} \ln \left( \frac{1 + x}{1 - x} \right) dx$. Substitute $t = \ln \left( \frac{1 + x}{1 - x} \right) = \ln(1 + x) - \ln(1 - x)$. Then $dt = \left( \frac{1}{1 + x} - \frac{-1}{1 - x} \right) dx = \left( \frac{1 - x + 1 + x}{1 - x^2} \right) dx = \frac{2}{1 - x^2} dx$. Thus,$\frac{dx}{1 - x^2} = \frac{1}{2} dt$. When $x = 0$,$t = \ln(1) = 0$. When $x = \frac{1}{2}$,$t = \ln \left( \frac{1 + 1/2}{1 - 1/2} \right) = \ln \left( \frac{3/2}{1/2} \right) = \ln 3$. Therefore,$I = \int\limits_0^{\ln 3} t \cdot \frac{1}{2} dt = \frac{1}{2} \left[ \frac{t^2}{2} \right]_0^{\ln 3} = \frac{1}{4} \ln^2 3$. Since $\ln^2(1/3) = (\ln 1 - \ln 3)^2 = (-\ln 3)^2 = \ln^2 3$,the result is $\frac{1}{4} \ln^2 3$ or $\frac{1}{4} \ln^2 \left( \frac{1}{3} \right)$.

$\int\limits_0^{\frac{1}{2}} \frac{1}{1 - x^2} \ln \left( \frac{1 + x}{1 - x} \right) dx$ is equal to :

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