evaluate the integral. (remember the constant of integration.) ∫x^(-2)(x^4 - 3/(2x^4))dx

Explanation:

Step1: Expand the integrand

First, expand $x^{-2}(x^{4}-\frac{3}{2x^{4}})$ using the distributive - property $a(b - c)=ab - ac$. We get $x^{-2}\cdot x^{4}-x^{-2}\cdot\frac{3}{2x^{4}}=x^{-2 + 4}-\frac{3}{2}x^{-2-4}=x^{2}-\frac{3}{2}x^{-6}$.

Step2: Integrate term - by - term

Use the power rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$). For $\int x^{2}dx$, with $n = 2$, we have $\frac{x^{2 + 1}}{2+1}=\frac{x^{3}}{3}$. For $\int-\frac{3}{2}x^{-6}dx$, with $n=-6$, we have $-\frac{3}{2}\cdot\frac{x^{-6 + 1}}{-6 + 1}=-\frac{3}{2}\cdot\frac{x^{-5}}{-5}=\frac{3}{10}x^{-5}$.

Answer:

$\frac{x^{3}}{3}+\frac{3}{10x^{5}}+C$