# Integrate the function$\sqrt{1-4x-x^2}$

$\begin{array}{1 1} \frac{(x+2) (\sqrt{1-4x-x^2})}{2}+\frac{5}{2} \sin^{-1}( \frac{x+2}{\sqrt 5})+c \\ \frac{(x+2) (\sqrt{1-4x-x^2})}{2}-\frac{5}{2} \cos^{-1}( \frac{x+2}{\sqrt 5})+c \\ \frac{(x-2) (\sqrt{1-4x-x^2})}{2}+\frac{5}{2} \sin^{-1}( \frac{x-2}{\sqrt 5})+c \\ \frac{(x-2) (\sqrt{1-4x-x^2})}{2}+\frac{5}{2} \cos^{-1}(\frac{x-2}{\sqrt 5})+c \end{array}$

Toolbox:
• $\int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac {a^2}{2}\sin^{-1}(\frac{x}{a})+c$
Given $I=\sqrt{1-4x-x^2}dx$

$(1-4x-x^2)=-(x^2+4x-1)$

This can split as $-[(x^2+4x+4)-4-1]$

$=-[(x+2)^2-(\sqrt 5)^2]$

$I=[(\sqrt 5)^2-(x+2)^2]$

Hence $I=\int \sqrt{(\sqrt 5)^2-(x+2)^2}dx$

This is of the form $\sqrt{a^2-x^2} dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}(\frac{x}{a})+c$

Here $a=\sqrt 5.x=(x+2)$

Hence$\int \sqrt{1-4x-x^2}=\frac{(x+2)}{2}\sqrt{1-4x-x^2}+\frac{5}{2} \sin^{-1}(\frac{x+2}{\sqrt 5})+c$