Integrate the function in Exercise: x+2 x^2+2x+3

Question

Integrate the function in Exercise:
$\frac{\text{x}+2}{\sqrt{\text{x}^2+2\text{x}+3}}$

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Answer

$\text{Let I}=\int\frac{\text{x}+2}{\sqrt{\text{x}^2+2\text{x}+3}}\text{ dx} \ \ \ \ ...\text{(i)}$
$\text{Let Linear}=\text{A}\frac{\text{d}}{\text{dx}}(\text{Quadratic})+\text{B}$
$\Rightarrow\ \ \text{x}+2=\text{A}\frac{\text{d}}{\text{dx}}(\text{x}^2+2\text{x}+3)+\text{B}$
$\Rightarrow\ \ \text{x}+2=\text{A}(2\text{x}+2)+\text{B}\ \ \ \ ...\text{(ii)}$
$\Rightarrow\ \ \text{x}+2=2\text{Ax}+2\text{A}+\text{B}$
Comparing coefficients of x,
$2\text{A}=1\Rightarrow\ \ \text{A}=\frac{1}{2}$
Comparing constants,
$2\text{A}+\text{B}=2$
On solving, we get
$\text{A}=\frac{1}{2}, \ \ \text{B}=1$
Putting the values of $A$ and $B$ in eq. $(ii),$
$\text{x}+2=\frac{1}{2}(2\text{x}+2)+1$
Putting this value of $x + 2$ in eq. $(i),$
$\text{I}=\int\frac{\frac{1}{2}(2\text{x}+2)+1}{\sqrt{\text{x}^2+2\text{x}+3}}\text{ dx}$
$\text{I}=\frac{1}{2}\int\frac{2\text{x}+2}{\sqrt{\text{x}^2+2\text{x}+3}}\text{ dx}+\int\frac{1}{\sqrt{\text{x}^2+2\text{x}+3}}\text{ dx}$
$\Rightarrow\ \ \text{I}=\frac{1}{2}\text{I}_1+\text{I}_2\ \ \ \ \ \ ...\text{(iii)}$
$\text{Now }\text{ I}_1=\int\frac{2\text{x}+2}{\sqrt{\text{x}^2+2\text{x}+3}}\text{ dx}$
$\text{Putting }\text{ x}^2+2\text{x}+3=\text{t}\ \ \Rightarrow\ \ \ 2\text{x}+2=\frac{\text{dt}}{\text{dx}}\ \ \Rightarrow\ \ \ (2\text{x}+2)\text{ dx}=\text{dt}$
$\therefore\ \ \text{I}_1=\int\frac{\text{dt}}{\sqrt{t}}=\int\text{t}^{\frac{-1}{2}}\text{ dt}=\frac{\text{t}^{\frac{1}{2}}}{\frac{1}{2}}$
$=2\sqrt{\text{t}}=2\sqrt{\text{x}^2+2\text{x}+3} \ \ \ \ ...\text{(iv)}$
$\text{Again I}_2=\int\frac{1}{\sqrt{\text{x}^2+2\text{x}+3}}\text{ dx}$
$=\int\frac{1}{\sqrt{\text{x}^{2}+2\text{x}+1+2}}=\int\frac{1}{\sqrt{(\text{x}+1)^2+(\sqrt{2})^2}}\text{ dx}$
$=\log\begin{vmatrix}\text{x}+1+\sqrt{(\text{x}+1)^2+(2)^2}\end{vmatrix}$
$=\log\begin{vmatrix}\text{x}+1+\sqrt{\text{x}^2+2\text{x}+3}\end{vmatrix} \ \ \ \ ...\text{(v)}$
Putting values of $I_1 $ and $I_2$ in eq. $(iii),$
$\text{I}=\sqrt{\text{x}^2+2\text{x}+3}+\log\begin{vmatrix}\text{x}+1+\sqrt{\text{x}^2+2\text{x}+3}\end{vmatrix}+\text{c}$

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