Find the indefinite integral of 1/x^2 - 1/(3 * sqrt(x)) with respect to x.

Step-by-step solution

1. 1

   Rewrite both terms as powers of $x$. The power rule cannot be applied until the roots and reciprocals are gone:

   $\frac{1}{x^2}=x^{-2},\qquad \frac{1}{3\sqrt x}=\frac13x^{-1/2}$

   This conversion is where the problem is actually decided; everything after it is mechanical.

2. 2

   Integrate $x^{-2}$. Adding one to the exponent gives $-1$:

   $\int x^{-2}dx=\frac{x^{-1}}{-1}=-\frac1x$

   The power rule applies because the new exponent is not $-1$; only $\int x^{-1}dx$ needs a logarithm instead.

3. 3

   Integrate $\tfrac13x^{-1/2}$. Adding one to $-\tfrac12$ gives $+\tfrac12$:

   $\int\frac13x^{-1/2}dx=\frac13\cdot\frac{x^{1/2}}{1/2}=\frac{2}{3}x^{1/2}=\frac23\sqrt x$

   Dividing by $\tfrac12$ doubles the coefficient, which is why the $\tfrac13$ becomes $\tfrac23$.

4. 4

   Combine with the minus sign and one constant.

   $\int\left(\frac{1}{x^2}-\frac{1}{3\sqrt x}\right)dx=-\frac1x-\frac23\sqrt x+C$

5. 5

   Differentiate to verify.

   $\frac{d}{dx}\left(-x^{-1}-\frac23x^{1/2}+C\right)=x^{-2}-\frac23\cdot\frac12x^{-1/2}=\frac{1}{x^2}-\frac{1}{3\sqrt x}\;\checkmark$

   Note the domain: the integrand needs $x>0$ for the square root, so the antiderivative is valid on $(0,\infty)$.