In order to find the integral of 1 x 2, we will use the power rule. In other words, we want to use
$$ \dfrac d{dx}x^n = nx^{n-1}$$
to findÂ
$$\displaystyle\int \dfrac{1}{x^2}dx$$
At first, it is not clear how the two are related. Recall, however, that another name for the integral is antiderivative. In other words, we wish to “undo” the derivative operation. When we apply the power rule to an antiderivative we get
$$\displaystyle\int \dfrac{1}{x^2}dx = \displaystyle\int x^{-2}dx = -1(x^{-1}) + C= -\frac1{x} +C$$
Whenever we take an antiderivative and are not sure of the result, we may check our answer by differentiation. In this case, we will try to differentiate
$$\dfrac d{dx}(-1(x^{-1}) + C) = 1(x^{-2}) = \frac{1}{x^2}$$
This is exactly what we were looking for, so the antiderivative was correct. In general, you can always check the antiderivative, also known as indefinite integral, by differentiating your answer. It is possible to generalize this method to antiderivatives of all powers:
$$\int x^n dx = \frac1{n+1} x^{n+1} + C$$
However, there is one value of $n$ for which this method would not work. This special value is $n = -1$. The integral of $x^{-1}$ is
$$\int\frac 1x dx = \ln |x| + C$$Â
It is important to note that the natural logarithm is of the absolute value of $x$, since the logarithm function is defined only for positive values of $x$. Other than this special case, antiderivatives of $x^n$ are very straightforward.
