Integration using partial fractions.

[Silvanoshei]

Need some help understanding the setup for....

\(\displaystyle ∫\frac{5x+7}{x^{2}+4x+4}\)

\(\displaystyle ∫\frac{5x+7}{(x+2)(x+2)}\)

Now if you set AB as....

\(\displaystyle \frac{A}{(x+2)}+\frac{B}{(x+2)}\)

Putting \(\displaystyle x=0\) is not going to help you out finding A and B. What's the trick here?

 

[pka]

Need some help understanding the setup for....

\(\displaystyle ∫\frac{5x+7}{x^{2}+4x+4}\)

Have a look at this webpage.

 

[stapel]

Need some help understanding the setup for....

\(\displaystyle ∫\frac{5x+7}{x^{2}+4x+4}\)

\(\displaystyle ∫\frac{5x+7}{(x+2)(x+2)}\)

Now if you set AB as....

\(\displaystyle \frac{A}{(x+2)}+\frac{B}{(x+2)}\)

Putting \(\displaystyle x=0\) is not going to help you out finding A and B. What's the trick here?

To learn how to set up and solve a decomposition into partial fractions, try here. In this particular case, you'd use the standard set-up:

. . . . .\(\displaystyle \dfrac{A}{x\, +\, 2}\, +\, \dfrac{B}{\left(x\, +\, 2\right)^2}\, =\, \dfrac{5x\, +\, 7}{\left(x\, +\, 2\right)^2}\)

Multiplying through by the common denominator gives us:

. . . . .\(\displaystyle A(x\, +\, 2)\, +\, B\, =\, 5x\, +\, 7\)

. . . . .\(\displaystyle (A)x\, +\, (2A\, +\, B)\, =\, (5)x\, +\, 7\)

Then \(\displaystyle A\, =\, 5\) and \(\displaystyle 2A\, +\, B\, =\, 7\). Solve the system to find the two fractions.

 

[lookagain]

Need some help understanding the setup for....

\(\displaystyle ∫\frac{5x+7}{x^{2}+4x+4}\)

\(\displaystyle ∫\frac{5x+7}{(x+2)(x+2)}\)

Don't forget the "dx" parts.

\(\displaystyle ∫\frac{5x+7}{x^{2}+4x+4}dx\)

\(\displaystyle ∫\frac{5x+7}{(x+2)(x+2)}dx\)