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