Question

Integrate the function: \(\frac{6x+7}{√(x-5)(x-4)}\)

Answer 1

\(\frac{6x+7}{√(x-5)(x-4)}\) =\(\frac{6x+7}{√x^2-9x+20}\)

Let 6x+7 = A\(\frac{d}{dx}\)(x²-9x+20)+B

⇒ 6x+7 = A(2x-9)+B

Equating the coefficients of x and constant term, we obtain
2A = 6 ⇒ A = 3
−9A + B = 7 ⇒ B = 34 
∴ 6x + 7 = 3 (2x − 9) + 34

∫\(\frac{6x+7}{√x^2-9x+20}\) = ∫\(\frac{3(2x-9)+34}{\sqrt{x^2-9x+20}}\) dx

=3 ∫\(\frac{(2x-9)}{\sqrt{x2-9x+20}}\)+ 34 ∫\(\frac{1}{\sqrt{x2-9x+20}}\) dx

Let I₁ = ∫\(\frac{(2x-9)}{\sqrt{x^2-9x+20}}\) and I₂ = ∫\(\frac{1}{\sqrt{x^2-9x+20}}\)dx

∴ ∫\(\frac{(6x+7)}{\sqrt{x^2-9x+20}}\) = 3I₁+34I₂                                         ...(1)

Then,
I₁ = ∫\(\frac{(2x-9)}{\sqrt{x^2-9x+20}}\) dx

Let x²-9x+20 = t
⇒ (2x-9)dx = dt
⇒I₁ = dt/√t
I₁ = 2√t
I₁ = 2√x²-9x+20                                                                 ...(2)
and I₂ = ∫\(\frac{1}{\sqrt{x^2-9x+20}}\)dx
x²-9x+20 can be written as x²-9x+20+\(\frac{81}{4}\)-\(\frac{81}{4}\)
Therefore, 
x²-9x+20+\(\frac{81}{4}\)-\(\frac{81}{4}\)
=(x-\(\frac{9}{2}\))²-1/4
=(x-\(\frac{9}{2}\))²-(1/2)²
⇒ I₂ = ∫1/√(x-\(\frac{9}{2}\))²-(1/2)² dx
 I₂ = log|(x-\(\frac{9}{2}\))+√x²-9x+20|                                                 ...(3)
Substituting equations (2) and (3) in (1), we obtain
∫6x+7/√x2-9x+20 dx = 3[2√x2-9x+20]+34 log[(x-\(\frac{9}{2}\))+√x2-9x+20]+C
=6√x2-9x+20 + 34 log[(x-\(\frac{9}{2}\))+√x2-9x+20]+C