# Evaluate : $\int \large\frac{6x+7}{\sqrt{(x-5)(x-4)}}$$dx ## 1 Answer Toolbox: • \int\large\frac{(px+q)}{\sqrt{ax^2+bx+c}}$$dx.$,where p,q,a,b,c are constants,we are to find real numbers such that $px+q=A\large\frac{d}{dx}$$(ax^2+bx+c)+B • Therefore px+q=A(2ax+b)+B. • To determine A and B we equate the coefficients from both sided,the coefficients of x and the constant terms.A and B are thus obtained • \int\large\frac{dx}{\sqrt{x^2-a^2}}$$=\log|x+\sqrt{x^2-a^2}|+c.$
Step 1:
$I=\int\large \frac{6x+7}{(x-5)(x-4)}$$dx=\int\large\frac{6x+7}{\sqrt{x^2-9x+20}}. We can write, 6x+7=A\large\frac{d}{dx}$$(x^2-9x+20)+B.$
$\;\;\;\quad=A(2x-9)+B.$
Now equating the coefficients we get,
6=2A$\Rightarrow A=3.$
$7=-9A+B\Rightarrow B=34.$
Hence A=3 and B=34.
Step 2:
$I=\int\large\frac{3(2x-9)+34}{\sqrt{x^2-9x+20}}$.
On separating we can write,
$I=3\int\large\frac{(2x-9)dx}{\sqrt{x^2-9x+20}}$$+34\int\frac{dx}{\sqrt{x^2-9x+20}}. Let x^2-9x+20=t. and x^2-9x+20=(x-\frac{9}{2})^2-\frac{1}{4}=(x-\large\frac{9}{2})^2-(\frac{1}{2})^2. On differentiating (2x-9)dx=dt. Therefore I=3\int\frac{dt}{\sqrt t}+34\int\frac{dx}{\sqrt{(x-\frac{9}{2}}})^2-(\large\frac{1}{2})^2. Step 3: On integrating we get, \;\;\;=3\begin{bmatrix}\large\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}\end{bmatrix}+34[\log|(x-9/2)+\sqrt{(x-5)(x-4)}|]. \;\;\;=3\times 2(\sqrt t)+34\log|(x-9/2)+\sqrt{(x-5)(x-4)}+c. Substituting for t we get, \int\large\frac{6x+7}{\sqrt{(x-5)(x-4)}}$$dx=6(\sqrt{x^2-9x+2})+34[\log|(x-9/2)+\sqrt{(x-5)(x-4)}|]+c.$