logo

Ask Questions, Get Answers

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

Integrate the functions\[\frac{5x-2}{1+2x+3x^2}\]

Can you answer this question?
 
 

1 Answer

0 votes
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)$\int\frac{dx}{\sqrt{x^2+a^2}}=\frac{1}{a}\tan^{-1}\big(\frac{x}{a}\big)+c.$
Given:$I=\int \frac{5x-2}{1+2x+3x^2}dx.$
 
Let $5x-2=A\frac{d}{dx}(1+2x+3x^2)+B.$
 
5x-2=A(2+6x)+B.
 
Equating the coefficients we get,
 
5=6A $\Rightarrow A=\frac{5}{6}.$
 
-2=2A+B.
 
$-2=2\big(\frac{5}{6}\big)+B\Rightarrow B=\frac{-11}{3}.$
 
Hence $I=\int\frac{5/6(2+6x)-11/3}{1+2x+3x^2}.$
 
On separating we can write,
 
$I=\frac{5}{6}\int\frac{2+6x}{1+2x+3x^2}-\frac{11}{3}\int\frac{dx}{1+2x+3x^2}.$
 
Put 1+2x+3x^2=t.$
 
On differentiating
 
(2+6x)dx=dt.
 
$1+2x+3x^2=3(x^2+\frac{2}{3}+\frac{1}{3})$.
 
$\qquad=3\big(x+\frac{1}{3}\big)^2+\big(\frac{\sqrt 2}{3}\big)^2$
 
On substituting for t and dt,
 
$I=\frac{5}{6}\int\frac{dt}{t}-\frac{11}{3}\int\frac{dx}{3(x+1/3)^2+\big(\frac{\sqrt 2}{3}}\big)^2$.
 
$\;\;\;=\frac{5}{6}\int\frac{dt}{t}-\frac{11}{9}\int\frac{dx}{(x+1/3)^2+\big(\frac{\sqrt 2}{3}}\big)^2$.
 
On integrating we get,
 
$\;\;\;=\frac{5}{6}log|t|-\frac{11}{9}\frac{3}{\sqrt 2}\tan^{-1}\big(\frac{x+1/3}{\sqrt 2/3}\big).$
 
Substituting for t we get,
 
$\;\;\;=\frac{5}{6}log(1+2x+3x^2)-\frac{11}{3\sqrt 2}\tan^{-1}\frac{(3x+1)}{\sqrt 2}$.

 

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