Question

find the integral.
int \frac { 1 } { x ^ { 9 } } d x
int \frac { 1 } { x ^ { 9 } } d x =

Explanation:

Step1: Rewrite the integrand

Rewrite \(\frac{1}{x^{9}}\) as \(x^{-9}\). So the integral becomes \(\int x^{-9}dx\).

Step2: Apply the power - rule for integration

The power - rule for integration is \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)). Here \(n=-9\).

$$
LATEXBLOCK0
$$

Step3: Simplify the result

\(\frac{x^{-8}}{-8}+C=-\frac{1}{8x^{8}}+C\)

Answer:

\(-\frac{1}{8x^{8}}+C\)