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