How do you integrate #(3x^2+2x)/((x+2)(x^2+4))# using partial fractions?

Question

Emma Aldridge · Accepted Answer

# int(3x^2+2x)/((x+2)(x^2+4)) dx= ln(x+2)+ ln(x^2+4)−arctan(x/2)+C #

The partial fraction decomposition will be of the form:

# (3x^2+2x)/((x+2)(x^2+4)) -= A/(x+2) + (Bx+C)/(x^2+4) # # :. (3x^2+2x)/((x+2)(x^2+4)) = (A(x^2+4) + (Bx+C)(x+2))/((x+2)(x^2+4)) # # :. 3x^2+2x = A(x^2+4) + (Bx+C)(x+2) #

We now need to find the three coefficients, #A#,#B# and #C#:

Put # x=-2=>12-4=A(4+4)+0 # # :. 8A=8 => A=1 #

Equating coefficients of #x^2 => 3=A+B# # :. B=3-1=2 #

Equating coefficients of constants # => 0=4A+2C # # :. 2C=-4 => C=-2 #

So, we have: # (3x^2+2x)/((x+2)(x^2+4)) = 1/(x+2) + (2x-2)/(x^2+4) #

And, therefore, # int(3x^2+2x)/((x+2)(x^2+4)) dx= int 1/(x+2) + (2x-2)/(x^2+4) dx# # :. int(3x^2+2x)/((x+2)(x^2+4)) dx= int 1/(x+2) dx+ int(2x-2)/(x^2+4) dx#

# :. int(3x^2+2x)/((x+2)(x^2+4)) dx= ln(x+2)+ ln(x^2+4)−arctan(x/2)+C#

I have omitted the derivation of the second interval, as the question is about partial fractions and not integration by substation, you can use https://tutor.hix.ai to get a full step by step.

Emilia Aguilar · Answer

undefined To integrate $\frac{3x^2+2x}{(x+2)(x^2+4)}$ using partial fractions, you would first express the fraction as a sum of simpler fractions. The denominator $ (x+2)(x^2+4) $ can be factored as $ (x+2)(x+2i)(x-2i) $. Then you would express the original fraction in terms of the sum of partial fractions as follows:

$$
\frac{3x^2+2x}{(x+2)(x^2+4)} = \frac{A}{x+2} + \frac{Bx + C}{x^2+4}
$$

By multiplying both sides by the denominator $(x+2)(x^2+4)$ and equating coefficients of like terms, you can solve for the constants $A$, $B$, and $C$. Once you find the values of $A$, $B$, and $C$, you can integrate each term separately. The integral of $\frac{A}{x+2}$ is $A\ln|x+2|$, and the integral of $\frac{Bx+C}{x^2+4}$ can be found using a substitution or other integration techniques.