∫1−x7x(1+x7)dx=al n|x|+bl n|x7+1|+c⇒(a, b)=

∫1−x7x(1+x7)dx=-∫x7-1x(1+x7)dx=-∫x7+1-2x(1+x7)dx =-∫1x+2∫1x1+x7dx =-log x+2∫x6x71+x7dx =-log x+27∫dttt+1 =-logx+27∫-1t+1+1t =-log x+27-logt+1+log t+C =-log x+27-logx7+1+log x7+C =-log x-27logx7+1+27.7 log x+C =log x-27log x7+1+C a=1, b=-27

∫1−x7x(1+x7)dx=al n|x|+bl n|x7+1|+c⇒(a, b)=

∫1−x7x(1+x7)dx=-∫x7-1x(1+x7)dx=-∫x7+1-2x(1+x7)dx =-∫1x+2∫1x1+x7dx =-log x+2∫x6x71+x7dx =-log x+27∫dttt+1 =-logx+27∫-1t+1+1t =-log x+27-logt+1+log t+C =-log x+27-logx7+1+log x7+C =-log x-27logx7+1+27.7 log x+C =log x-27log x7+1+C a=1, b=-27

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$\int \frac{1 - x^{7}}{x (1 + x^{7})} d x = a l n | x | + b l n | x^{7} + 1 | + c \Rightarrow (a, b) =$

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$(1, \frac{2}{7})$

$(1, \frac{- 7}{2})$

$(1, \frac{- 2}{7})$

$(2, \frac{- 2}{7})$

answer is C.

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Detailed Solution

$\int \frac{1 - x^{7}}{x (1 + x^{7})} d x$

$= - \int \frac{x^{7} - 1}{x (1 + x^{7})} d x$

$= - \int \frac{(x^{7} + 1) - 2}{x (1 + x^{7})} d x = - \int \frac{1}{x} + 2 \int \frac{1}{x (1 + x^{7})} d x = - \log x + 2 \int \frac{x^{6}}{x^{7} (1 + x^{7})} d x = - \log x + \frac{2}{7} \int \frac{d t}{t (t + 1)} = - \log x + \frac{2}{7} \int [\frac{- 1}{t + 1} + \frac{1}{t}] = - \log x + \frac{2}{7} [- \log |t + 1| + \log t] + C = - \log x + \frac{2}{7} (- \log (x^{7} + 1) + \log x^{7}) + C = - \log x - \frac{2}{7} \log (x^{7} + 1) + \frac{2}{7} . 7 \log x + C = \log x - \frac{2}{7} \log (x^{7} + 1) + C a = 1, b = \frac{- 2}{7}$