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

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Homework Help Overview

The discussion revolves around the integration of the function (x+2)/(x^2-2x+3) dx, with participants exploring various methods and approaches to solve the integral. The subject area is calculus, specifically focusing on integration techniques.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of partial fraction decomposition and consider different ways to manipulate the integrand. There is a suggestion to rewrite the integrand in a different form to facilitate integration. Some participants express concerns about specific terms complicating the integration process.

Discussion Status

The discussion is active, with participants offering hints and suggestions without providing direct solutions. There is acknowledgment of the challenges faced in the integration process, and some participants are exploring multiple interpretations of the problem.

Contextual Notes

Participants note the constraints of not providing complete solutions and the need to explore various methods within the context of homework help. There is an emphasis on understanding the structure of the integrand and the implications of its components.

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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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