# Evaluate the definite integral$\int\limits_2^3\frac{xdx}{x^2+1}$

$\begin{array}{1 1}\large \frac{\log 2}{2} \\ \large \frac{\log 2}{4} \\ \large \frac{\log 2}{3} \\\large \frac{3\log 2}{2} \end{array}$

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$
edited Apr 11, 2016 by meena.p