Find the value of $\int_1^2\frac{3x-5}{x^3}~dx$. I want to find the value of 
$$\int_1^2\frac{3x-5}{x^3}~dx$$
I'm not sure how to proceed, due to the difference in degree between $3x-5$ and $x^3$. I've tried the substitution $u = 3x-5$, giving the equality
$$\int_1^2\frac{3x-5}{x^3}~dx = 9\int_{-2}^1 \frac{u}{(u-5)^3}~du$$
However, this doesn't solve the problem I have of the difference in degree between the numerator and the denominator. I thought that perhaps partial fractions would make a difference, but the quadratic I had to solve was untenable. 

Notice that 
\begin{equation}
 \frac{3x - 5}{x^3}
 =
 \frac{3x}{x^3}
 -
 \frac{5}{x^3}
 =
 \frac{3}{x^2}
 -
 \frac{5}{x^3}
 =
 3 x^{-2} - 5x^{-3}
\end{equation}
So the integration becomes
\begin{equation}
 3 \int x^{-2} - 5 \int x^{-3}
 =
 3 \frac{x^{-3}}{-3}
 -
 5
 \frac{x^{-4}}{-4}
\end{equation}
Can you take it from here ? 
