Integral Homework Help: Solving (x+2)/(x^2-2x+3) dx | Step-by-Step Guide

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The discussion revolves around integrating the function (x+2)/(x^2-2x+3) dx. Initial attempts involved using partial fraction decomposition but faced challenges with the integration. A suggestion was made to rewrite the integrand to facilitate the solution, specifically by breaking it into two fractions. Ultimately, the user successfully solved the integral with the provided hints. The thread emphasizes collaborative problem-solving in integral calculus.
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Homework Statement


Integrate

Homework Equations



I started with (2x^2+3)/(x^3-2x^2+3x) dx

The Attempt at a Solution



Got dx/x + (x+2)/(x^2-2x+3) dx using the A/x + (Bx+C)/(x^2-2x+3) method.

But I can't seem to solve (x+2)/(x^2-2x+3) dx

Possible solutions are welcome.
 
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grimireaper said:
I started with (2x^2+3)/(x^3-2x^2+3x) dx

and got dx/x + (x+2)/(x^2-2x+3) dx using the A/x + (Bx+C)/(x^2-2x+3) method.

But I can't seem to solve (x+2)/(x^2-2x+3) dx

Possible solutions are welcome.

Welcome to PF. You won't get solutions here, but you may get hints. Try this:$$
\frac {x+2}{x^2-2x+3} = \frac{(x-1)+3}{(x-1)^2+2}$$and see if that gives you any ideas.
 
I thought of that but the +2 at the bottom ruins it, because splitting up x-1 and 3 into 2 integrals won't help, as I can't cancel anything out
 
grimireaper said:
I thought of that but the +2 at the bottom ruins it, because splitting up x-1 and 3 into 2 integrals won't help, as I can't cancel anything out

But you can break it up into two fractions and use different methods on them.
 
$$2x^2+3=(x^2-x)+(3x)+(x^2-2x+3)=\frac{x}{2}\left(\frac{x^3-2x^2+3x}{x}\right)^\prime+3\frac{x^3-2x^2+3x}{x^2-2x+3}+\frac{x^3-2x^2+3x}{x}$$
 
Got it solved. Thank you for the help.
 
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