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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the definite integral\[\int\limits_0^2\frac{6x+3}{x^2+4}dx\]

$\begin{array}{1 1}3 \log 2+\large\frac{3\pi}{8} \\\log 2+\large\frac{\pi}{8} \\ 5 \log 2+\large\frac{5\pi}{6} \\ 5 \log 2+\large\frac{3\pi}{8} \end{array} $

Can you answer this question?
 
 

1 Answer

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Toolbox:
  • (i)$\int \limits_a^b f(x)dx=F(b)-F(a)$
  • (ii)$ \int \large\frac{1}{x^2+a^2}dx=\frac{1}{a} \tan^{-1}(x/a)+c$
Given $ I=\int\limits_0^2\large\frac{6x+3}{x^2+4}dx$
 
On seperating the terms
 
$ I=\int\limits_0^2\large\frac{6x}{x^2+4}dx+\int\limits_0^2\large\frac{3}{x^2+4}dx$
 
Consider $ I_1=\int\limits_0^2\large\frac{6x}{x^2+4}dx$
 
Let $x^2+4=t$
 
On differentiating w.r.t.x
 
$ 2xdx=dt$
 
Therefore $=>6xdx=3dt$
 
Substituting for t and dt,
 
$\int _0^2 \large\frac{3dt}{t}$
 
The limits change when we substitute for t
 
When x =  0, t= 4
when x =   2, t= 8
 
On integrating we get
 
$I_1=[3 log t]_4^8------(1)$
 
Consider $ I_2=\int\limits_0^2\large\frac{3}{x^2+4}dx$
 
$=3\int\limits_0^2 \large\frac{dx}{x^2+4}dx$
 
$=\bigg[3 \frac{1}{2}\tan^{-1}(x/2)\bigg]_0^2-------(2)$
 
Now $I=I_1+I_2$
 
Therefore $I=\bigg[3log t\bigg]_4^8+\bigg[\frac{3}{2} \tan^{-1}x/2\bigg]_0^2$
 
$I=\bigg[log t\bigg]_4^8+\bigg[\frac{3}{2} \tan^{-1}x/2\bigg]_0^2$
 
On applying limits,
 
$=(log 8^3-log 4^3)+\frac{3}{2}\bigg[(\tan^{-1}(2/2)-\tan 0)\bigg]$
 
$=log \large\frac{8^3}{4^3}+\frac{3}{2}\tan ^{-1}-0$
 
$(log a-log b=log a/b)$
 
we know $\tan^{-1}(1)=\frac{\pi}{4}$
 
Hence $log 2^3+\frac{3}{2}\frac{\pi}{4}$
 
$=3 log 2+\frac{3}{2}\frac{\pi}{4}$
 
$\int\limits_0^2\large\frac{6x+3}{x^2+4}dx=3 log 2+\frac{3\pi}{8}$

 

 

answered Feb 12, 2013 by meena.p
edited Aug 1, 2013 by vijayalakshmi_ramakrishnans
 
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