Question

$$ int \frac { 1 } { 2 x ^ { 2 } } d x $$

Explanation:

Step1: Simplify the integrand

The integrand is \(\frac{1}{2x^{2}}=\frac{1}{2}x^{- 2}\). So the integral becomes \(\int\frac{1}{2}x^{-2}dx\).

Step2: Use the power rule for integration

The power rule for integration is \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)). For \(n=-2\), we have \(\int\frac{1}{2}x^{-2}dx=\frac{1}{2}\int x^{-2}dx\).
Substituting \(n =-2\) into the power - rule formula: \(\frac{1}{2}\times\frac{x^{-2 + 1}}{-2+1}+C\).

Step3: Simplify the result

\(\frac{1}{2}\times\frac{x^{-1}}{-1}+C=-\frac{1}{2x}+C\).

Answer:

\(-\frac{1}{2x}+C\)