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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} $

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  • (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
While substituting for t,then the limits also change
Hence when x=2
when x=3
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,
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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