Answer
$-\displaystyle \frac{10}{x^{0.1}} -\ln|x| +C$
Work Step by Step
$\displaystyle \int(\frac{1}{x^{1.1}}-\frac{1}{x})dx=$
The integral of a sum is the sum of the integrals.
Also, write $\displaystyle \frac{1}{x^{n}}=x^{-n}$
=$\displaystyle \int(x^{-1.1}-x^{-1})dx$
$=\displaystyle \int x^{-1.1}dx-\int x^{-1}dx$
for $n=-1, \displaystyle \qquad \int x^{n}dx=\ln|x|+C$
for $n\neq-1,\quad \displaystyle \int x^{n}dx=\frac{x^{n+1}}{n+1}+C$
$=\displaystyle \frac{x^{-0.1}}{-0.1} -\ln|x| +C$
= $-\displaystyle \frac{10}{x^{0.1}} -\ln|x| +C$
