If ∫1xx5−1x5+1dx=Alog⁡x+Blog⁡x5−1+Clog⁡x5+1 then (A,B,C)=

I=∫1xx5−1x5+1dx I=15∫5x4x5x5−1x5+1dx Put x5=t⇒5x4dx=dt I=15∫dttt-1t+1= 1tt-1t+1=Pt+Qt-1+Rt+1 1=Pt-1t+1+Qtt+1+Rtt-1 t=0⇒P=-1, t=1⇒Q=12 t=-1⇒R=12 ⇒I=15∫dttt−1t+1=15∫-1t+12t-1+12t+1dt ⇒I=-15logt+110logt-1+110logt+1+C ⇒I=-logx+110logx5-1+110logx5+1+C A=-1,B=110,C=110

If ∫1xx5−1x5+1dx=Alog⁡x+Blog⁡x5−1+Clog⁡x5+1 then (A,B,C)=

I=∫1xx5−1x5+1dx I=15∫5x4x5x5−1x5+1dx Put x5=t⇒5x4dx=dt I=15∫dttt-1t+1= 1tt-1t+1=Pt+Qt-1+Rt+1 1=Pt-1t+1+Qtt+1+Rtt-1 t=0⇒P=-1, t=1⇒Q=12 t=-1⇒R=12 ⇒I=15∫dttt−1t+1=15∫-1t+12t-1+12t+1dt ⇒I=-15logt+110logt-1+110logt+1+C ⇒I=-logx+110logx5-1+110logx5+1+C A=-1,B=110,C=110

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$If \int \frac{1}{x (x^{5} - 1) (x^{5} + 1)} d x = A \log x + B \log (x^{5} - 1) + C \log (x^{5} + 1) then (A, B, C) =$

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$(- 1, \frac{1}{10}, \frac{1}{10})$

$(- \frac{5}{4}, \frac{1}{8}, - \frac{1}{8})$

$(- \frac{5}{4}, - \frac{1}{8}, \frac{1}{8})$

$(1, \frac{1}{10}, - \frac{1}{10})$

answer is A.

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Detailed Solution

$I = \int \frac{1}{x (x^{5} - 1) (x^{5} + 1)} d x I = \frac{1}{5} \int \frac{5 x^{4}}{x^{5} (x^{5} - 1) (x^{5} + 1)} d x Put x^{5} = t \Rightarrow 5 x^{4} d x = d t$

$I = \frac{1}{5} \int \frac{d t}{t (t - 1) (t + 1)} = \frac{1}{t (t - 1) (t + 1)} = \frac{P}{t} + \frac{Q}{t - 1} + \frac{R}{t + 1} 1 = P (t - 1) (t + 1) + Q t (t + 1) + R t (t - 1) t = 0 \Rightarrow P = - 1, t = 1 \Rightarrow Q = \frac{1}{2} t = - 1 \Rightarrow R = \frac{1}{2}$
$\Rightarrow I = \frac{1}{5} \int \frac{d t}{t (t - 1) (t + 1)} = \frac{1}{5} \int (\frac{- 1}{t} + \frac{1}{2 (t - 1)} + \frac{1}{2 (t + 1)}) d t \Rightarrow I = \frac{- 1}{5} \log t + \frac{1}{10} \log (t - 1) + \frac{1}{10} \log (t + 1) + C \Rightarrow I = - \log x + \frac{1}{10} \log (x^{5} - 1) + \frac{1}{10} \log (x^{5} + 1) + C A = - 1, B = \frac{1}{10}, C = \frac{1}{10}$