Science:Math Exam Resources/Courses/MATH105/April 2016/Question 06/Solution 1

By the partial fraction decomposition, we integrand can be written as

$\frac{1}{x^{2016}-x}=\frac{1}{x(x^{2015}-1)}=\frac{x^{2014}}{x^{2015}-1}-\frac{1}{x}.$

Taking indefinite integral, we obtain

$\int \frac{dx}{x^{2016}-x}=\int \frac{x^{2014}}{x^{2015}-1}dx-\int \frac{1}{x}dx=\int \frac{x^{2014}}{x^{2015}-1}dx-\ln x+C.$

Using substitution $u=x^{2015}-1$ (and hence $du=2015x^{2014}dx$), the first integral can be computed as

$\int \frac{x^{2014}}{x^{2015}-1}dx=\frac{1}{2015}\int \frac{1}{u}du=\frac{1}{2015}\ln u+C=\frac{1}{2015}\ln (x^{2015}-1)+C.$

Therefore, the answer is $\int \frac{dx}{x^{2016}-x}=\frac{1}{2015}\ln (x^{2015}-1)-\ln x+C.$