∫ ln x - 1 x + 1 x 2 - 1 dx is equal to, 1 2 ln x - 1 x + 1 2 + C, 1 2 ln x + 1 x - 1 2 + C, 1 4 ln x - 1 x + 1 2 + C, 1 4 ln x + 1 x - 1 + C

I = ∫ ln x - 1 x + 1 x 2 - 1 dx , Let t = ln x - 1 x + 1 ⇒ dt dx = x + 1 x - 1 x + 1 - x - 1 x + 1 2 = 2 x 2 - 1 ⇒ dx x 2 - 1 = dt 2 ⇒ I = 1 2 ∫ tdt = 1 4 t 2 + C = 1 4 ln x - 1 x + 1 2 + C

∫ ln x - 1 x + 1 x 2 - 1 dx is equal to

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Question: Answered & Verified by Expert

\int \frac{ln (\frac{x - 1}{x + 1})}{x^{2} - 1} dx

is equal to

MathematicsIndefinite IntegrationJEE Main

Options:

A $\frac{1}{2} {(ln (\frac{x - 1}{x + 1}))}^{2} + C$
B $\frac{1}{2} {(ln (\frac{x + 1}{x - 1}))}^{2} + C$
C $\frac{1}{4} {(ln (\frac{x - 1}{x + 1}))}^{2} + C$
D $\frac{1}{4} (ln (\frac{x + 1}{x - 1})) + C$

Solution:

2496 Upvotes Verified Answer

The correct answer is:

\frac{1}{4} {(ln (\frac{x - 1}{x + 1}))}^{2} + C

I = \int \frac{ln (\frac{x - 1}{x + 1})}{x^{2} - 1} dx

,
Let

t = ln (\frac{x - 1}{x + 1})

\Rightarrow \frac{dt}{dx} = \frac{x + 1}{x - 1} \{\frac{x + 1 - (x - 1)}{{(x + 1)}^{2}}\} = \frac{2}{(x^{2} - 1)}

\Rightarrow \frac{dx}{x^{2} - 1} = \frac{dt}{2}

\Rightarrow I = \frac{1}{2} \int tdt = \frac{1}{4} t^{2} + C = \frac{1}{4} {(ln (\frac{x - 1}{x + 1}))}^{2} + C

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