Question

Solve: $\int \frac{1}{x(x^4-1)}dx$

• A. $-\frac{1}{4}\ln \left|\frac{x^4-1}{x^4}\right|+c$
• B. $\frac{1}{4}\ln \left|\frac{x^4-1}{x^4}\right|+c$
• C. $-\frac{1}{4}\ln \left|\frac{x^2-1}{x^2}\right|+c$
• D. $\frac{1}{4}\ln \left|\frac{x^2-1}{x^2}\right|+c$

Answer 1

The correct option is B $\frac{1}{4}\ln \left|\frac{x^4-1}{x^4}\right|+c$

Take $I=\int \frac{1}{x(x^4-1)}dx$

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

$1=A(x+1)(x-1)(x^2+1)+Bx(x-1)(x^2+1)+Cx(x+1)(x^2+1)+Dx(x+1)(x-1)$

For $x=1,A=-1$

For $x=1,C=\frac{1}{4}$

For $x=-1,B=\frac{1}{4}$

For $x=2,D=\frac{1}{4}$

Therefore $\int \frac{1}{x(x^4-1)}dx=-\int \frac{1}{x}dx+\frac{1}{4}\int \frac{dx}{x+1}+\frac{1}{4}\int \frac{dx}{x-1}+\frac{1}{4}\int \frac{dx}{x^2+1}$

$=-\ln |x|+\frac{1}{4}\ln |(x+1)|+\frac{1}{4}\ln |(x-1)|+\frac{1}{4}\ln |(x^2+1)|+c$

$=\frac{1}{4}\ln \left|\frac{x^4-1}{x^4}\right|+c$