Solution

Let $I=\int \sqrt{4-x^{2}} d x=\int \sqrt{(2)^{2}-(x)^{2}} d x$

It is known that, $\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}} \frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+C$

$ \begin{aligned} \therefore I & =\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}+C \\ & =\frac{x}{2} \sqrt{4-x^{2}}+2 \sin ^{-1} \frac{x}{2}+C \end{aligned} $

Solution

Let $I=\int \sqrt{1-4 x^{2}} d x=\int \sqrt{(1)^{2}-(2 x)^{2}} d x$

Let $2 x=t \Rightarrow 2 d x=d t$

$\therefore I=\frac{1}{2} \int \sqrt{(1)^{2}-(t)^{2}} d t$

It is known that, $\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+C$

$ \begin{aligned} \Rightarrow I & =\frac{1}{2}[\frac{t}{2} \sqrt{1-t^{2}}+\frac{1}{2} \sin ^{-1} t]+C \\ & =\frac{t}{4} \sqrt{1-t^{2}}+\frac{1}{4} \sin ^{-1} t+C \\ & =\frac{2 x}{4} \sqrt{1-4 x^{2}}+\frac{1}{4} \sin ^{-1} 2 x+C \\ & =\frac{x}{2} \sqrt{1-4 x^{2}}+\frac{1}{4} \sin ^{-1} 2 x+C \end{aligned} $

Solution

Let $I=\int \sqrt{x^{2}+4 x+6} d x$

$ \begin{aligned} & =\int \sqrt{x^{2}+4 x+4+2} d x \\ & =\int \sqrt{(x^{2}+4 x+4)+2} d x \\ & =\int \sqrt{(x+2)^{2}+(\sqrt{2})^{2}} d x \end{aligned} $

It is known that, $\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log |x+\sqrt{x^{2}+a^{2}}|+C$

$ \begin{aligned} \therefore I & =\frac{(x+2)}{2} \sqrt{x^{2}+4 x+6}+\frac{2}{2} \log |(x+2)+\sqrt{x^{2}+4 x+6}|+C \\ & =\frac{(x+2)}{2} \sqrt{x^{2}+4 x+6}+\log |(x+2)+\sqrt{x^{2}+4 x+6}|+C \end{aligned} $

Solution

Let $I=\int \sqrt{x^{2}+4 x+1} d x$

$ \begin{aligned} & =\int \sqrt{(x^{2}+4 x+4)-3} d x \\ & =\int \sqrt{(x+2)^{2}-(\sqrt{3})^{2}} d x \end{aligned} $

It is known that, $\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log |x+\sqrt{x^{2}-a^{2}}|+C$

$\therefore I=\frac{(x+2)}{2} \sqrt{x^{2}+4 x+1}-\frac{3}{2} \log |(x+2)+\sqrt{x^{2}+4 x+1}|+C$

Solution

Let $I=\int \sqrt{1-4 x-x^{2}} d x$

$ \begin{aligned} & =\int \sqrt{1-(x^{2}+4 x+4-4)} d x \\ & =\int \sqrt{1+4-(x+2)^{2}} d x \\ & =\int \sqrt{(\sqrt{5})^{2}-(x+2)^{2}} d x \end{aligned} $

It is known that, $\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+C$

$\therefore I=\frac{(x+2)}{2} \sqrt{1-4 x-x^{2}}+\frac{5}{2} \sin ^{-1}(\frac{x+2}{\sqrt{5}})+C$

Solution

Let $I=\int \sqrt{x^{2}+4 x-5} d x$

$ \begin{aligned} & =\int \sqrt{(x^{2}+4 x+4)-9} d x \\ & =\int \sqrt{(x+2)^{2}-(3)^{2}} d x \end{aligned} $

It is known that, $\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log |x+\sqrt{x^{2}-a^{2}}|+C$

$\therefore I=\frac{(x+2)}{2} \sqrt{x^{2}+4 x-5}-\frac{9}{2} \log |(x+2)+\sqrt{x^{2}+4 x-5}|+C$

Solution

Let $I=\int \sqrt{1+3 x-x^{2}} d x$

$ \begin{aligned} & =\int \sqrt{1-(x^{2}-3 x+\frac{9}{4}-\frac{9}{4})} d x \\ & =\int \sqrt{(1+\frac{9}{4})-(x-\frac{3}{2})^{2}} d x \\ & =\int \sqrt{(\frac{\sqrt{13}}{2})^{2}-(x-\frac{3}{2})^{2}} d x \end{aligned} $

It is known that, $\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+C$

$ \begin{aligned} \therefore I & =\frac{x-\frac{3}{2}}{2} \sqrt{1+3 x-x^{2}}+\frac{13}{4 \times 2} \sin ^{-1}(\frac{x-\frac{3}{2}}{\frac{\sqrt{13}}{2}})+C \\ & =\frac{2 x-3}{4} \sqrt{1+3 x-x^{2}}+\frac{13}{8} \sin ^{-1}(\frac{2 x-3}{\sqrt{13}})+C \end{aligned} $

Solution

Let $I=\int \sqrt{x^{2}+3 x} d x$

$ \begin{aligned} & =\int \sqrt{x^{2}+3 x+\frac{9}{4}-\frac{9}{4}} d x \\ & =\int \sqrt{(x+\frac{3}{2})^{2}-(\frac{3}{2})^{2}} d x \end{aligned} $

It is known that, $\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log |x+\sqrt{x^{2}-a^{2}}|+C$

$ \begin{aligned} \therefore I & =\frac{(x+\frac{3}{2})}{2} \sqrt{x^{2}+3 x}-\frac{9}{2} \log |(x+\frac{3}{2})+\sqrt{x^{2}+3 x}|+C \\ & =\frac{(2 x+3)}{4} \sqrt{x^{2}+3 x}-\frac{9}{8} \log |(x+\frac{3}{2})+\sqrt{x^{2}+3 x}|+C \end{aligned} $

Solution

Let $I=\int \sqrt{1+\frac{x^{2}}{9}} d x=\frac{1}{3} \int \sqrt{9+x^{2}} d x=\frac{1}{3} \int \sqrt{(3)^{2}+x^{2}} d x$

It is known that, $\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log |x+\sqrt{x^{2}+a^{2}}|+C$

$ \begin{aligned} \therefore I & =\frac{1}{3}[\frac{x}{2} \sqrt{x^{2}+9}+\frac{9}{2} \log |x+\sqrt{x^{2}+9}|]+C \\ & =\frac{x}{6} \sqrt{x^{2}+9}+\frac{3}{2} \log |x+\sqrt{x^{2}+9}|+C \end{aligned} $

Solution

It is known that, $\int \sqrt{a^{2}+x^{2}} d x=\frac{x}{2} \sqrt{a^{2}+x^{2}}+\frac{a^{2}}{2} \log |x+\sqrt{x^{2}+a^{2}}|+C$

$ \therefore \int \sqrt{1+x^{2}} d x=\frac{x}{2} \sqrt{1+x^{2}}+\frac{1}{2} \log |x+\sqrt{1+x^{2}}|+C $

Hence, the correct Answer is A.

Solution

Let $I=\int \sqrt{x^{2}-8 x+7} d x$

$ \begin{aligned} & =\int \sqrt{(x^{2}-8 x+16)-9} d x \\ & =\int \sqrt{(x-4)^{2}-(3)^{2}} d x \end{aligned} $

It is known that, $\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log |x+\sqrt{x^{2}-a^{2}}|+C$

$ \therefore I=\frac{(x-4)}{2} \sqrt{x^{2}-8 x+7}-\frac{9}{2} \log |(x-4)+\sqrt{x^{2}-8 x+7}|+C $

Hence, the correct Answer is D.