# Evaluate the definite integral$\int\limits_0^2\frac{6x+3}{x^2+4}dx$

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

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 $log 2^3+\frac{3}{2}\frac{\pi}{4}$

$=3 log 2+\frac{3}{2}\frac{\pi}{4}$

$\int\limits_0^2\large\frac{6x+3}{x^2+4}dx=3 log 2+\frac{3\pi}{8}$

edited Aug 1, 2013