Calculus 2 for Engineers Question

Question

∫ (x^2-4x+4)dx/((12+4x-x^2)^(3/2))

 

Any help with this integral would be much appreciated. Is partial fraction decomposition necessary here?

 

Thank you!

Answer 1

Actually partial fraction decomposition isn't necessary for this problem! Just some tricky rearranging and trigonometric substitution:

 

First, the numerator is:

x² - 4x + 4 = (x - 2)²

 

and the denominator is:

12 + 4x - x² = 16 - (x² - 4x + 4x) = 16 - (x-2)²  <-- and take this to the 3/2 power

 

We see a common (x - 2), so let's set u = x - 2, and since it's just a translation, du = dx. Putting all that in, we get:

 

∫ [ u² du / (16 - u²)^3/2 ]

 

Here's where the trigsub comes in. Let's set u = 4cosθ, so that du = -4sinθ dθ, and that

(16 - u²)^3/2 = (16 - 16cos²θ)^3/2 = 64sin³θ. I hope I didn't lose you in how I got those values.

 

The integrand is then:

 

[ u² du / (16 - u²)^3/2 ] = [ (16cos²θ)(-4sinθ) / 64sin³θ ] dθ = - cot²θ dθ = (1 - csc²θ) dθ

 

And there we have it! The integral is simply:

 

∫ (1 - csc²θ) dθ = θ + cotθ

 

Or, if we remember that u = x - 2 = 4cosθ  =>  x = 2 + 4cosθ  =>  θ = cos^-1 [ (x-2) / 4 ]

                                                                                             =>  cotθ = (x-2) / √[16 - (x-2)²]

 

The answer is cos^-1 [ (x-2) / 4 ] + (x-2) / √[16 - (x-2)²]

 

Hope this helps! Ask me if more clarification on some steps are needed