Integration using Algebraic Manipulation Trying to solve 
$$\int^1_2\frac{x-4}{x^2}dx$$
All I have so far is re-writing it into two integrals which yields the following: 
$$\int^1_2\frac{x}{x^2} \, dx - \int^1_2\frac{-4}{x^2} \,dx$$
Then $\frac{x}{x^2}$ can be re-written as $\frac{1}{x}$ which yields: 
$$\int^1_2\frac{1}{x} \,dx - \int^1_2\frac{-4}{x^2} \, dx$$
I know that $\displaystyle \int\frac{1}{x} \, dx =\ln(x)$ but I'm not sure where to go from here. Could someone perhaps help?

For one, you can rewrite $$\begin{align}\int^1_2\frac{x-4}{x^2}dx = \int^2_1\frac{4-x}{x^2}dx \\ =4\int^2_1\frac{1}{x^2}dx - \int^2_1\frac{x}{x^2}dx \\ =4\int^2_1\frac{1}{x^2}dx - \int^2_1\frac{1}{x}dx \end{align}$$ and you are correct that $\int \frac{1}{x}dx = \ln|x|+C$. As for the other term to be integrated, consider this. For any constants $a,n,C \in \Bbb{R}$ we know by the power rule that $$\frac{d}{dx}ax^n+C = anx^{n-1} \quad \implies \quad \int anx^{n-1}dx = ax^n+C$$ Can you pattern match your integral of $\int\frac{4}{x^2}dx = \int 4x^{-2}dx$ to the general form of $\int anx^{n-1}dx$ for some $n$ and $a$?
