Answer:
[tex]x^2+4x -\frac{1}{2} lnx + \frac{2}{13} ln(x-2) + C[/tex]
Step-by-step explanation:
Given the integrand [tex]\int\limits{\dfrac{2x^3 - 2x + 1}{x^2-2x} } \, dx[/tex], before evaluating the integral function, we will need to simplify the function first by applying long division as shown in the attachment.
Hence the partial form of the function [tex]\dfrac{2x^3 - 2x + 1}{x^2-2x} } = 2x+4 + \frac{6x+1}{x^2-2x}[/tex]
Integrating its partial sum
[tex]\int\limits \dfrac{2x^3 - 2x + 1}{x^2-2x} }dx = \int\limits (2x+4 + \frac{6x+1}{x^2-2x})\ dx\\\\= \int\limits {2x} \, dx + \int\limits {4} \, dx + \int\limits {\frac{6x+1}{x^2-2x} \, dx\\ = \frac{2x^2}{2}+4x -\frac{1}{2} \int\limits{\frac{1}{x} } \, dx + \frac{2}{13} \int\limits{\frac{1}{x-2} } \, dx[/tex]
[tex]= \frac{2x^2}{2}+4x -\frac{1}{2} lnx + \frac{2}{13} ln(x-2) + C[/tex]
[tex]= x^2+4x -\frac{1}{2} lnx + \frac{2}{13} ln(x-2) + C[/tex]
NB: Find the partial sum calculation also in the attachment.
