# How do you integrate int 1/x^3dx?

Nov 5, 2016

Rewrite as $\int {x}^{-} 3 \mathrm{dx}$ and take the anti-derivative.

#### Explanation:

We first need to recognize that $\frac{1}{x} ^ 3$ is equivalent to ${x}^{-} 3$.

Once we get that far, the problem becomes quite simple to solve.

Do be careful, however, as we are dealing with a negative exponent, so when we add one to the power as we take the anti-derivative, the magnitude of the power will decrease. This also means that our constant will have to be negative as well, since the x term in the integral is positive. Taking the anti-derivative, we get:

$- \frac{1}{2} \cdot {x}^{-} 2 + C$

This is equivalent to $- \frac{1}{2 {x}^{2}} + C$

As usual, you can check this answer by taking the derivative, which gives you ${x}^{-} 3$.