Indefinite Integration of $\frac {2}{x^2}-\frac 1{2x^3}$

Question

The question is to find the integral of this:

$$\int\left(\frac 2{x^2}-\frac 1{2x^3}\right)\, dx$$

I get basic integration but I'm not sure about how to hand the $2x^3$. Any ideas?

Answer 1

Everyone should know, not only $$\int x^{n}\,\mathrm dx=\frac{x^{n+1}}{n+1}\quad (n\in \mathbf R,\; n\ne- 1),$$ but also the ‘denominator form’: $$\int\frac{\mathrm dx}{x^n}=-\frac1{(n-1)\mkern1mux^{n-1}}\quad(n\in \mathbf R,\; n\ne 1). $$ Knowing that, you obtain readily $$\int\biggl(\frac 2{x^2}-\frac 1{2x^3}\biggr)\,\mathrm dx=2\int\frac {\mathrm dx}{x^2}-\frac12\int\frac{\mathrm dx}{x^3}=-\frac2x+\frac12\frac1{2\mkern1mu x^2}=\frac{1-2x}{4x^2}.$$

Answer 2

Hint: write your Integrand in the form $$2x^{-2}-\frac{1}{2}x^{-3}$$ and use that $$\int x^ndx=\frac{x^{n+1}}{n+1}+C$$ if $$n\neq -1$$