∫11−x2log1+x1−xdx=

I=∫11-x2log 1+x1-xdx Put log 1+x1-x=t ⇒11+x1-x×(1-x)(1)-(1+x)(-1)(1-x)2dx=dt ⇒ 1 (1-x+1+x)(1+x) (1-x) dx=dt ⇒ dx1-x2=dt2 I=∫ tdt2 =t24+c =14log 1+x1-x2+c

∫11−x2log1+x1−xdx=

I=∫11-x2log 1+x1-xdx Put log 1+x1-x=t ⇒11+x1-x×(1-x)(1)-(1+x)(-1)(1-x)2dx=dt ⇒ 1 (1-x+1+x)(1+x) (1-x) dx=dt ⇒ dx1-x2=dt2 I=∫ tdt2 =t24+c =14log 1+x1-x2+c

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$\int \frac{1}{1 - x^{2}} \log (\frac{1 + x}{1 - x}) d x =$

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$\frac{1}{2} \log {(\frac{1 + x}{1 - x})}^{2}$

$\frac{1}{4} {(\log (\frac{1 + x}{1 - x}))}^{2} + c$

$\frac{1}{3} \log (\frac{1 + x}{1 - x}) + c$

${(\frac{1}{2} \log (\frac{1 + x}{1 - x}))}^{3} + c$

answer is B.

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Detailed Solution

$I = \int \frac{1}{1 - x^{2}} \log (\frac{1 + x}{1 - x}) dx Put \log (\frac{1 + x}{1 - x}) = t \Rightarrow \frac{1}{\frac{1 + x}{1 - x}} \times \frac{(1 - x) (1) - (1 + x) (- 1)}{{(1 - x)}^{2}} dx = d t \Rightarrow \frac{1 (1 - x + 1 + x)}{(1 + x) (1 - x)} dx = dt \Rightarrow \frac{dx}{1 - x^{2}} = \frac{dt}{2} I = \int t \frac{dt}{2} = \frac{t^{2}}{4} + c = \frac{1}{4} {[\log (\frac{1 + x}{1 - x})]}^{2} + c$