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

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

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## 1 Answer

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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,

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

answered Feb 4, 2013
edited Feb 4, 2013

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