Question

The integral of $\int \frac{dx}{x^{13}(x^6-1)}$ is:

• A. $\frac{1}{6}\ln \left|\frac{x^6-1}{x^6}\right|+c$
• B. $\frac{1}{6}[\ln \left|\frac{x^6-1}{x^6}\right|+x^{-6}+\frac{1}{2}{x}^{-12}]+c$
• C. $\frac{1}{6}[\ln \left|\frac{x^6-1}{x^6}\right|+x^{-6}+x^{-12}]+c$
• D. $\frac{1}{6}[\ln \left|\frac{x^6-1}{x^6}\right|+x^{-6}+\frac{1}{4}{x}^{-12}]+c$

Answer 1

The correct option is B $\frac{1}{6}[\ln \left|\frac{x^6-1}{x^6}\right|+x^{-6}+\frac{1}{2}{x}^{-12}]+c$

$I=\int \frac{dx}{x^{13}(x^6-1)}=\int \frac{x^{5}dx}{x^{18}(x^6-1)}$
Put $x^6=t\Rightarrow 6x^{5}dx=dt$
$\Rightarrow I=\int \frac{dt}{6t^3(t-1)}=\frac{1}{6}\int \frac{dt}{t^3(t-1)}$
Using partial fractions, we have:
$\frac{1}{t^3(t-1)}=\frac{A}{t-1}+\frac{B{t}^2+Ct+D}{t^3}$
$\Rightarrow \frac{1}{t^3(t-1)}=\frac{A{t}^3+B{t}^3+C{t}^2-B{t}^2+Dt-Ct-D}{t^3(t-1)}$
On comparing the coefficents, we have:
$D=C=B=-1;A=1$
$\therefore I=\frac{1}{6}\int (\frac{1}{t-1}-\frac{t^2+t+1}{t^3})dt$
$\Rightarrow I=\frac{1}{6}\int (\frac{1}{t-1}-\frac{1}{t}-\frac{1}{t^2}-\frac{1}{t^3})dt$
$\Rightarrow I=\frac{1}{6}[\ln \left(\frac{t-1}{t}\right)-\frac{t^{-1}}{-1}-\frac{t^{-2}}{-2}]+c$
$\Rightarrow I=\frac{1}{6}[\ln \left|\frac{x^6-1}{x^6}\right|+x^{-6}+\frac{1}{2}{x}^{-12}]+c$