I have


Factorising the denominator I have


From there I split the top term into two parts to make it easier to integrate

$\int{\frac{2x+1}{(x+2)(x+2)}}dx$ = $\int{\frac{A}{(x+2)}+\frac{B}{(x+2)}}dx$



$2x+1 = A(x+2) +B(x+2)$

This is where I would normally use a substitution method to eliminate either the A term or B term by letting x = something like -2, which would get rid of the A and usually leave me with the B term to solve. However since they are the same I'm not sure what to do.

I've been told to try evaluate the co-efficients, but am not sure how.


You want to try a split like

$$\frac{2 x+1}{(x+2)^2} = \frac{A}{x+2} + \frac{B x}{(x+2)^2}$$

Then $A+B=2$ and $2 A=1$. The decomposition is then

$$\frac{2 x+1}{(x+2)^2} = \frac12 \frac1{x+2} + \frac{3}{2} \frac{x}{(x+2)^2}$$

  • $\begingroup$ Is it $Bx$? I doubt $\endgroup$ – Semsem Apr 6 '14 at 9:06
  • $\begingroup$ @semsem: you doubt what? Check the work, it is correct. I could have done just $B$ as well. The reason either way works is that one may add and subtract $2$ in the numerator of the second fraction to change the representation. $\endgroup$ – Ron Gordon Apr 6 '14 at 9:12
  • $\begingroup$ Ok, you are right $\endgroup$ – Semsem Apr 6 '14 at 9:13



  • $\begingroup$ This is the best what can be done. $\endgroup$ – kmitov Apr 6 '14 at 8:47

All you need to do is to solve this with respect to polynomials:



$A+B=2 \rightarrow B=2-A$


$2A+4-2A=1\rightarrow 4=1$

This is contradiction! You have made an mistake in step where you split the term into two fractions, you should have done it like this:


and then proceed like usual.





$\int \frac{2}{x+2}+\frac{-3}{(x+2)^2}dx=2ln(x+2)+\frac{3}{x+2}+C$


In this case we a linear repeated factor so we split it like $$\frac{A}{(x+2)}+\frac{B}{(x+2)^2}=\frac{A(x+2)+B}{(x+2)^2}$$ and hence $$2x+1=A(x+2)+B=Ax+2A+B$$ then by equating coefficients $$A=2,\\2A+B=1\implies B=-3$$

  • $\begingroup$ How did you get A=2 from that? $\endgroup$ – user88720 Apr 9 '14 at 6:14
  • $\begingroup$ @user88720 i edited it $\endgroup$ – Semsem Apr 9 '14 at 7:07

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