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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the definite integral\[\int\limits_2^3\frac{xdx}{x^2+1}\]

$\begin{array}{1 1}\large \frac{\log 2}{2} \\ \large \frac{\log 2}{4} \\ \large \frac{\log 2}{3} \\\large \frac{3\log 2}{2} \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)Methods of substitution:$ \int f(x)dx.\; Let f(x)=t \; then f'(x)dx=dt.Hence\; \int f(x)dx=\int x.dt$
Given $\int\limits_2^3\large\frac{xdx}{x^2+1}$
 
This can be solved by the method of substitution:
 
Let $ x^2+1=t$
 
On differentiating with respect to x
 
$2xdx=dt$
 
$=>xdx=\large\frac{dt}{2}$
 
While substituting for t,then the limits also change
 
Hence when x=2
 
$t=2^2+1=5$
 
when x=3
 
$t=3^2+1=10$
 
Hence here a=5 and b=10
 
The integral function becomes
 
$\int \limits_5^{10} \large\frac{dt/2}{t}=\frac{1}{2}\int \limits_5^{10} \large\frac{dt}{t}$
 
On integrating,
 
$I=\bigg[\large\frac{1}{2} log (t)\bigg]^{10}_5$
 
On applying limits,
 
$=\frac{1}{2}[log(10)-log(5)]$
 
But $log a -log b=log(a/b),$similarly
 
$ I=\large\frac{1}{2}$$[log \frac{10}{5}]=\frac{1}{2}log 2$
 
$\int \limits_2^3 \large\frac{xdx}{x^2+1}=\frac{1}{2}$$log 2$
answered Feb 11, 2013 by meena.p
edited Apr 11 by meena.p
 
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