Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Integrals
0 votes

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} $

Can you answer this question?

1 Answer

0 votes
  • $(i)\int\frac{(px+q)}{\sqrt{ax^2+bx+c}}dx.$,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,
Now equating the coefficients we get,
6=2A$\Rightarrow A=3.$
$7=-9A+B\Rightarrow B=34.$
Hence A=3 and B=34.
On separating we can write,
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
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,
$\;\;\;=3\times 2(\sqrt t)+34log|(x-9/2)+\sqrt{(x-5)(x-4)}+c.$
Substituting for t we get,


answered Feb 4, 2013 by sreemathi.v
edited Feb 4, 2013 by sreemathi.v
Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App