how do you integrate 1/ (16 + x^2)^2?

Answer 1

∫1/(16 + x²)² dx

First, make a trigonometric substitution - let θ = arctan (x/4), so x = 4 tan θ, and dx = 4 sec² θ dθ. Then we have:

∫1/(16 + 16 tan² θ)² * 4 sec² θ dθ

Using the identity tan² θ + 1 = sec² θ:

∫4 sec² θ/(16 sec² θ)² dθ

Simplifying:

∫4 sec² θ/(256 sec⁴ θ) dθ
1/64 ∫cos² θ dθ

Using the identity cos² θ = (cos (2θ) + 1)/2:

1/128 ∫cos (2θ) + 1 dθ

Integrating:

sin (2θ)/256 + θ/128 + C

Now we need to find sin (2θ) in terms of x. Let y=sin (2θ). Then we have:

y=sin (2θ)
y²=sin² (2θ)=1-cos² (2θ)
y²=1-(2 cos² θ - 1)² = 1 - 4 cos⁴ θ + 4 cos² θ - 1
y²=4/sec² θ - 4/(sec² θ)²
y²=4/(1+tan² θ) - 4/(1+tan² θ)²
y²=4/(1+(x/4)²) - 4/(1+(x/4)²)²
y²=64/(16+x²) - 1024/(16+x²)²
y²=(1024 + 64x²)/(16+x²)² - 1024/(16+x²)²
y²=64x²/(16+x²)²
y=±8x/(16+x²)

Now we just need to figure out the correct sign of y. If x≥0, the θ=arctan (x/4) will be in the range [0, π/2), and so 2θ∈[0, π), so y=sin (2θ)≥0. Conversely, if x<0, then θ=arctan (x/4) will be in the range (-π/2, 0), so 2θ∈(-π, 0), so y=sin(2θ)<0. Thus in either case, the sign of y is the same as the sign of x, which, since 8/(16+x²) is always positive, is the same as the sign of 8x/(16+x²). Thus, y=8x/(16+x²). Substituting this into our expression for the integral and simplifying:

sin (2θ)/256 + θ/128 + C
8x/(16+x²) / 256 + arctan (x/4)/128 + C
x/(32(16+x²)) + arctan (x/4)/128 + C
x/(512+32x²) + arctan (x/4)/128 + C

And we are done.