int {frac{1}{{{x^2} - 1}}} ,ln left( {frac{{x | vedclass

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(D) Let $I = \int \frac{1}{x^2 - 1} \ln \left( \frac{x - 1}{x + 1} \right) dx$. We know that $\frac{1}{x^2 - 1} = \frac{1}{(x-1)(x+1)}$. Let $t = \ln

Vedclass · Accepted Answer

Both $(B)$ and $(C)$

Vedclass · Answer

(D) Let $I = \int \frac{1}{x^2 - 1} \ln \left( \frac{x - 1}{x + 1} \right) dx$. We know that $\frac{1}{x^2 - 1} = \frac{1}{(x-1)(x+1)}$. Let $t = \ln \left( \frac{x - 1}{x + 1} \right) = \ln(x-1) - \ln(x+1)$. Differentiating with respect to $x$: $dt = \left( \frac{1}{x-1} - \frac{1}{x+1} \right) dx = \left( \frac{(x+1) - (x-1)}{(x-1)(x+1)} \right) dx = \frac{2}{x^2 - 1} dx$. Thus,$\frac{1}{x^2 - 1} dx = \frac{1}{2} dt$. Substituting these into the integral: $I = \int t \cdot \frac{1}{2} dt = \frac{1}{2} \cdot \frac{t^2}{2} + C = \frac{1}{4} t^2 + C$. Substituting $t$ back: $I = \frac{1}{4} \ln^2 \left( \frac{x - 1}{x + 1} \right) + C$. Since $\ln \left( \frac{x + 1}{x - 1} \right) = - \ln \left( \frac{x - 1}{x + 1} \right)$,squaring it gives the same result: $\ln^2 \left( \frac{x + 1}{x - 1} \right) = \ln^2 \left( \frac{x - 1}{x + 1} \right)$. Therefore,both $(B)$ and $(C)$ are correct.

$\int {\frac{1}{{{x^2} - 1}}} \,\ln \left( {\frac{{x - 1}}{{x + 1}}} \right)dx$ equals :

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