Answer
a) $\dfrac{1}{3}\ln |x|+C$
b) $\dfrac{2}{5}\ln |x|+C$
c) $x+\dfrac{4}{3}\ln |x|+\dfrac{1}{x}+C$
Work Step by Step
a) The anti-derivative is:
$\dfrac{1}{3x}=\dfrac{1}{3}\ln |x|+C$
b) The anti-derivative is:
$\dfrac{2}{5}=\dfrac{2}{5}\ln |x|+C$
c) The anti-derivative is:
$1+\dfrac{4}{3x}-\dfrac{x^{-2+1}}{-2+1}=x+\dfrac{4}{3}\ln |x|+\dfrac{1}{x}+C$
