Browse Questions

# Integrate the function$\sqrt{x^2+4x+6}$

$\begin{array}{1 1} \large \frac{(x+2)(\sqrt{x^2+4x+6})}{2}+\log \mid (x+2)+\sqrt{x^2+4x+6} \mid +c \\\large \frac{(x+2)(\sqrt{x^2+4x+6})}{2}+\log \mid (x+2)-\sqrt{x^2+4x+6} \mid +c \\\large \frac{(x+2)(\sqrt{2x+4})}{2}+\log \mid (x+2)+\sqrt{x^2+4x+6} \mid +c \\\large \frac{(x+2)(\sqrt{2x+4})}{2}+\log \mid (x+2)+\sqrt{2x+4} \mid +c\end{array}$

Toolbox:
• $\int \sqrt{x^2+a^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac {a^2}{2}log|x+\sqrt{x^2+a^2}|+c$
Given $I=\sqrt{x^2+4x+6}dx$

we can split $I=\sqrt{x^2+4x+6}\;as\;\sqrt{(x^2+4x+4)-4+6}$

$\sqrt{(x+2)^2-4+6}=\sqrt{(x+2)^2+2}$

Hence $I=\sqrt{(x+2)^2+(\sqrt 2})^2$

This is of the form $\sqrt{x^2+a^2}=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}log|(x+2)+\sqrt{x^2+a^2}|+c$

Here x=(x+2) and $a=\sqrt 2.$

Therefore $I=\frac{(x+2)}{2}\sqrt{x^2+4x+6}+\frac{(\sqrt 2)^2}{2} log|x+\sqrt{x^2+4x+6}|+c$

$I=\frac{(x+2)}{2}\sqrt{x^2+4x+6}+log|(x+2)+\sqrt{x^2+4x+6}|+c$