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BaLLatris [955]
3 years ago
7

Perform long division on the​ integrand, write the proper fraction as a sum of partial​ fractions, and then evaluate the integra

l.
2x3 - 2x + 1/x2 - 2x dx

Mathematics
1 answer:
worty [1.4K]3 years ago
5 0

Answer:

x^2+4x -\frac{1}{2}  lnx  + \frac{2}{13}  ln(x-2) + C

Step-by-step explanation:

Given the integrand \int\limits{\dfrac{2x^3 - 2x + 1}{x^2-2x} } \, dx, 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 \dfrac{2x^3 - 2x + 1}{x^2-2x} } = 2x+4 + \frac{6x+1}{x^2-2x}

Integrating its partial sum

\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

=  \frac{2x^2}{2}+4x -\frac{1}{2}  lnx  + \frac{2}{13}  ln(x-2) + C

= x^2+4x -\frac{1}{2}  lnx  + \frac{2}{13}  ln(x-2) + C

<em>NB: Find the partial sum calculation also in the attachment. </em>

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