Question

$\int \frac{1}{x^{13}(x^6-1)}dx$

Answer 1

We have,

$\int \frac{1}{x^{13}(x^6-1)}dx$

On multiplying and dividing $x^5$

Then,

$\int \frac{x^5}{x^{18}(x^6-1)}dx$

$\Rightarrow \int \frac{x^5}{{\left(x^6\right)}^3(x^6-1)}dx$

Let,

$t=x^6$

$dt=6x^{5}dx$

$\frac{1}{5}dt=x^{5}dx$

Now,

$\int \frac{1}{6t^3(t-1)}dt$

By application of partial dfraction

Then,

$\int \frac{1}{6t^3(t-1)}dt=\int \frac{1}{6(t-1)}dy-\int \frac{1}{6t}dy-\int \frac{1}{6t^2}dy-\int \frac{1}{6t^3}dy$

$\int \frac{1}{6t^3(t-1)}dt=\frac{1}{6}\log (t-1)-\frac{1}{6}\log t+\frac{1}{6t}+\frac{1}{12t^2}$

Substitution of t, gives us

$\int \frac{1}{x^{13}(x^6-1)}dx=\frac{1}{6}\log (x^6-1)-\frac{1}{6}\log x^6+\frac{1}{6x^6}+\frac{1}{12{x^1}^2}$

Hence, this is the answer.