If ∫1xx−1x+1dx= ln f(x)+g(x)+c, f(1)=1,g(1)=0 then

∫1xx−1x+1dx=∫x−1xx2−1dx=∫(1x2−1− 1xx2−1)dx

If ∫1xx−1x+1dx= ln f(x)+g(x)+c, f(1)=1,g(1)=0 then

∫1xx−1x+1dx=∫x−1xx2−1dx=∫(1x2−1− 1xx2−1)dx

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If $\int \frac{1}{x} \sqrt{\frac{x - 1}{x + 1}} d x = \ln f (x) + g (x) + c, f (1) = 1, g (1) = 0$ then

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$g (\sqrt{2}) = \frac{π}{3}$

$f^{'} (\sqrt{2}) = \sqrt{2} + 1$

$f^{'} (1) = 0$

$g (\sqrt{2}) = \frac{π}{6}$

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$\begin{array}{l} \int \frac{1}{x} \sqrt{\frac{x - 1}{x + 1}} d x \\ = \int \frac{x - 1}{x \sqrt{x^{2} - 1}} d x \\ = \int (\frac{1}{\sqrt{x^{2} - 1}} - \frac{1}{x \sqrt{x^{2} - 1}}) d x \end{array}$

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If ∫1xx−1x+1dx= ln f(x)+g(x)+c, f(1)=1,g(1)=0 then

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If $\int \frac{1}{x} \sqrt{\frac{x - 1}{x + 1}} d x = \ln f (x) + g (x) + c, f (1) = 1, g (1) = 0$ then