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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the definite integral\[\int\limits_1^2\frac{5x^2}{x^2+4x+3}\]

$\begin{array}{1 1} 5- \large \frac{5}{2}[9 \log(5/4)-\log(3/2)] \\ 5- \large \frac{5}{2}[7 \log(5/4)+\log(3/2)] \\5+ \large \frac{5}{2}[9 \log(5/4)+\log(3/2)]\\ 5- \large \frac{3}{2}[7 \log(5/4)-\log(5/2)] \end{array} $

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1 Answer

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Toolbox:
  • (i)$\int \limits_a^bf(x)dx=F(b)-F(a)$
  • (ii)If the given rational function is improper in natture,then we can divide and split it into a proper rational function and then resolve it into partial fraction of the given form.
  • (iii)$\int \frac{dx}{(a+x)}=log(a+x)+c$
Given $\int\limits_1^2\large\frac{5x^2}{x^2+4x+3}$
 
The given function is an improper rational function. Hence on dividing we get
 
$\int\limits_1^2\large\frac{5x^2}{x^2+4x+3}=5-\frac{20x+15}{x^2+4x+3}$
 
$\large\frac{20x+15}{x^2+4x+3}$ can be resolved as $ \large\frac{A}{(x+3)}+\frac{B}{(x+1)}$
 
Became $(x^2+4x+3)$ can be factorised as $(x+3)(x+1)$
 
Therefore $20x+15=A(x+1)+B(x+3)$
 
Equating the coefficient of x,
 
$ 20=A+B\;------(1)$
 
Equating the constant term,
 
$15=A+3B\;------(2)$
 
On solving equation (1) and(2),
 
$A+3B=15$
$A+\;B=20$
___________
$\qquad 2B=-5$
 
Therefore $B=-\frac{5}{2}$
 
Substituting the value of B in equ(1)
 
$A-\frac{5}{2}=20$
 
$=>a=\frac{45}{2}$
 
Hence $A=\frac{45}{2}\;and\;B=-\frac{5}{2}$
 
Now substituting for A and B we get
 
$\large\frac{20x+15}{x^2+4x+3}=\frac{42}{2}\bigg(\frac{1}{x+3}\bigg)-\frac{5}{2}\bigg(\frac{1}{x+1}\bigg)$
 
On integtating we get
 
$\int \limits_1^2 \large\frac{5x^2}{x^2+4x+3}=\int \limits_1^2 5dx -\bigg[\int \limits_1^2 \frac{45}{2}\bigg(\frac{1}{x+3}\bigg)dx-\int
 
\limits_1^2 \frac{5}{2}\bigg(\frac{1}{x+1}\bigg)dx\bigg]$
 
$=\bigg[5x\bigg]_1^2-\large\frac{45}{2}[log(x+3)]_1^2+\frac{5}{2}[log (x+1)]_1^2$
 
On applying limits,
 
$[5 \times 2-5 \times 1]-\frac{45}{2}[log(x+3)-log(1+3)]-\frac{5}{2}[log(2+1)-log(1+1)]$
 
But we know $log(a)-log(b)=log(a/b)$
 
Therefore $ \int \limits_1^2 \large\frac{5x^2}{x^2+4x+3}=5-\frac{45}{2}[log(5/4)-log(3/2)]$
 
$ \int \limits_1^2 \large\frac{5x^2}{x^2+4x+3}=5-\frac{5}{2}[9log(5/4)-log(3/2)]$

 

answered Feb 11, 2013 by meena.p
 
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