How do you use partial fractions to find the integral #int (x^2+x+3)/(x^4+6x^2+9)dx#?

Question

Aurora Alvarado · Accepted Answer

See the answer below:

Scarlett Allison · Answer

#1/sqrt(3)tan^(-1)(x/sqrt(3))-1/(2(x^2+3))+C#

Noticing that the denominator is a perfect square, we see there is no need to use partial fractions.

#(x^2+x+3)/(x^4+6x^2+9)=(x^2+3+x)/(x^2+3)^2=cancel(x^2+3)/(x^2+3)^cancel(2)+x/(x^2+3)^2# #=1/(x^2+3)+x/(x^2+3)^2# The first term is a standard integral of the form #1/(x^2+a^2)# with #a=sqrt(3)#, yielding #(1/sqrt(3))tan^(-1)(x/sqrt(3))#.

The second term is of the form where the the numerator is the derivative of the contents of the brackets in the denominator, save for a factor of #2#, and so can be integrated at sight. (Alternatively substitute #u=x^2+3#, #dx/(du)=1/(2x)# to give #1/2int u^(-2)du#.)

In general, #d/(dx)(1/(f(x)))=-(f prime(x))/(f(x))^2# so #int (f prime(x))/(f(x))^2dx=-1/(f(x))#. In this case, #f(x)=x^2+3# and #f prime(x)=2x#.

Luke Alvarez · Answer

undefined First, factor the denominator: $ x^4 + 6x^2 + 9 = (x^2 + 3)^2 $. Then, express the integrand as partial fractions:

$$
\frac{x^2 + x + 3}{x^4 + 6x^2 + 9} = \frac{Ax + B}{x^2 + 3} + \frac{Cx + D}{(x^2 + 3)^2}
$$

Multiply both sides by the denominator $ x^4 + 6x^2 + 9 = (x^2 + 3)^2 $ to clear the fractions:

$$
x^2 + x + 3 = (Ax + B)(x^2 + 3) + (Cx + D)
$$

Expand and collect like terms:

$$
x^2 + x + 3 = Ax^3 + 3Ax + Bx^2 + 3B + Cx + D
$$

Now, equate coefficients for corresponding powers of $ x $:

For $ x^3 $: $ A = 0 $  
For $ x^2 $: $ B = 1 $  
For $ x $: $ 3A + C = 1 $  
For the constant term: $ 3B + D = 3 $

Solving these equations gives $ A = 0, B = 1, C = -3, D = 6 $. Now, rewrite the integrand using these partial fractions:

$$
\frac{x^2 + x + 3}{x^4 + 6x^2 + 9} = \frac{x}{x^2 + 3} - \frac{3x + 6}{(x^2 + 3)^2}
$$

Now integrate term by term:

$$
\int \frac{x}{x^2 + 3} \, dx - \int \frac{3x + 6}{(x^2 + 3)^2} \, dx
$$

For the first integral, use a substitution $ u = x^2 + 3 $ to get $ \frac{1}{2} \ln|x^2 + 3| + C_1 $.

For the second integral, use a substitution $ u = x^2 + 3 $ and apply integration by parts to get $ -\frac{1}{2(x^2 + 3)} - \frac{3}{2} \ln|x^2 + 3| + C_2 $.

Combine the results to get the final integral:

$$
\frac{1}{2} \ln|x^2 + 3| - \frac{1}{2(x^2 + 3)} - \frac{3}{2} \ln|x^2 + 3| + C
$$

Simplify to obtain the answer:

$$
\frac{1}{2} \ln|x^2 + 3| - \frac{1}{2(x^2 + 3)} - \frac{3}{2} \ln|x^2 + 3| + C = -\frac{1}{2} \ln|x^2 + 3| - \frac{1}{2(x^2 + 3)} + C
$$