Integrating partial fractions

Question

I have

$\int{\frac{2x+1}{x^2+4x+4}}dx$

Factorising the denominator I have

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

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

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

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

Therefore

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

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.

Answer 1

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

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

Answer 2

Hint

$$\int{\frac{2x+1\color{red}{+3-3}}{x^2+4x+4}}dx=2\ln|x+2|+\frac{3}{x+2}+C$$

Answer 3

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

$2x+1=Ax+2A+Bx+2B$

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

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

$2A+2B=1$

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

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:

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

and then proceed like usual.

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

$A=2$

$2A+B=1$

$B=-3$

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

Answer 4

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 $$A=2,\\2A+B=1\implies B=-3$$