Integrate the functions\[\frac{6x+7}{\sqrt{(x-5)(x-4)}}\]

Question

$\begin{array}{1 1} 6(\sqrt{x^2-9x+2})+34[\log|(x-9/2)+\sqrt{(x-5)(x-4)}|]+c \\ 6(\sqrt{x^2-9x+2})-34[\log|(x-9/2)+\sqrt{(x-5)(x-4)}|]+c \\ 6(\sqrt{x^2-9x+2})-17[\log|(x-9/2)-\sqrt{(x-5)(x-4)}|]+c \\ 6(\sqrt{x^2-9x+2} -17 [\log|(x+9/2)-\sqrt{(x-5)(x-4)}|]+c. \end{array} $

Answer 1

Toolbox:

• $(i)\int\frac{(px+q)}{\sqrt{ax^2+bx+c}}dx.$,where p,q,a,b,c are constants,we are to find real numbers such that \[px+q=A\frac{d}{dx}(ax^2+bx+c)+B.\]
• Therefore $px+q=A(2ax+b)+B.$
• To determine A and B we equate the coefficients from both sided,the coefficients of x and the constant terms.A and B are thus obtained and hence the integral is reduced to one of the known forms:
• (ii)$\int\frac{dx}{\sqrt{x^2-a^2}}=log|x+\sqrt{x^2-a^2}|+c.$

Given:$I=\int \frac{6x+7}{(x-5)(x-4)}dx=\int\frac{6x+7}{\sqrt{x^2-9x+20}}.$

 

We can write,

 

6x+7=$A\frac{d}{dx}(x^2-9x+20)+B.$

 

$\;\;\;=A(2x-9)+B.$

 

Now equating the coefficients we get,

 

6=2A$\Rightarrow A=3.$

 

$7=-9A+B\Rightarrow B=34.$

 

Hence A=3 and B=34.

 

$I=\int\frac{3(2x-9)+34}{\sqrt{x^2-9x+20}}$.

 

On separating we can write,

 

$I=3\int\frac{(2x-9)dx}{\sqrt{x^2-9x+20}}+34\int\frac{dx}{\sqrt{x^2-9x+20}}.$

 

Let $x^2-9x+20=t.$ and $x^2-9x+20=(x-\frac{9}{2})^2-\frac{1}{4}=(x-\frac{9}{2})^2-(\frac{1}{2})^2$.

 

On differentiating

 

$(2x-9)dx=dt.$

 

Therefore $I=3\int\frac{dt}{\sqrt t}+34\int\frac{dx}{\sqrt{(x-\frac{9}{2}}})^2-(\frac{1}{2})^2.$

 

On integrating we get,

 

$\;\;\;=3\begin{bmatrix}\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}\end{bmatrix}+34[log|(x-9/2)+\sqrt{(x-5)(x-4)}|]$.

 

$\;\;\;=3\times 2(\sqrt t)+34log|(x-9/2)+\sqrt{(x-5)(x-4)}+c.$

 

Substituting for t we get,

 

$\int\frac{6x+7}{\sqrt{(x-5)(x-4)}}dx=6(\sqrt{x^2-9x+2})+34[log|(x-9/2)+\sqrt{(x-5)(x-4)}|]+c.$