How do I integrate (2x+1)/(x^2+16)^2 dx?

Question

I'm suppose to use partial fractions, and I'm given the hint: "Each denominator contains a repeated irreducible quadratic factor".

Answer 1

Again, I disagree. The partial fraction decomposition will not produce rational expressions that can be directly integrated.

 

Try it. You will get:

 

 Ax+B             cx + d                       2x + 1

--------     +   ----------------     =   -------------

x^2 + 16        (x^2+16)^2             (x^2+16)^2

 

(Ax+B)(x^2+16) + (cx+d) = 2x + 1

 

Ax^3 + Bx^2 + 16AX + 16B + Cx +d = 2x + 1

 

a=0, b=0 --->  c=2 and d = 1

 

--------------------------------------------------------

 

Instead break the fraction into two pieces:

 

(2x)/(x^2 + 16) and 1/(x^2+16)

 

The first can be integrated with the substitution U = x^2 + 16.

Then dU = 2x dx

 

The second is integrated using the reduction formula:

 

 integral  1/(ax^2+b)^n  =   (2n-3)/( (2b)(n-1))  integral  1/(ax^2+b)^n-1 + x/((2b)(n-1)(ax^2+b)^(n-1))

 

with a=1 b=16 and n=2

 

After using this formula, you will get integral of 1/(x^2 +16) which integrates to arctan(x/4)

 

The final answer is (x-32)/(32x^2+512) + arctan(x/4)/128 + k

You can find the reduction formula in the table of integrals in your calculus book or
google them. Wikipedia has a nice listing.