Question

If $\int \frac{1}{x(x^5-1)(x^5+1)}dx=A\ln |x|+B\ln |x^5-1|+C\ln |x^5+1|+D,$ then which of the following is/ are correct
(where $A,B,C$ are fixed constants and $D$ is constant of integration)

• A. $B-C=0$
• B. $2A+30B=1$
• C. $A-B+C=3$
• D. $\frac{A}{B+C}=-5$

Answer 1

Answer: (A), (B), (D)

$\frac{A}{B+C}=-5$

Let
$I=\int \frac{1}{x(x^5-1)(x^5+1)}dx$
$\Rightarrow I=\int \frac{1}{x(x^{10}-1)}dx$
$\Rightarrow I=\int \frac{1}{x^{11}(1-x^{-10})}dx$
Put $1-x^{-10}=t$
$\Rightarrow 10x^{-11}dx=dt$
$\therefore I=\frac{1}{10}\int \frac{dt}{t}$
$\Rightarrow I=\frac{1}{10}\ln |t|+D$
$\Rightarrow I=\frac{1}{10}\ln |1-x^{-10}|+D$
$\Rightarrow I=\frac{1}{10}(\ln |x^{10}-1|-\ln x^{10})+D$
$\Rightarrow I=-\ln x+\frac{1}{10}\ln |x^5-1|+\frac{1}{10}\ln |x^5+1|+D$
$\therefore A=-1,B=\frac{1}{10}$ and $C=\frac{1}{10}$