Question:
$\int \sqrt{1+x^{2}} d x$ is equal to
A. $\frac{x}{2} \sqrt{1+x^{2}}+\frac{1}{2} \log \left|x+\sqrt{1+x^{2}}\right|+\mathrm{C}$
B. $\frac{2}{3}\left(1+x^{2}\right)^{\frac{2}{3}}+\mathrm{C}$
C. $\frac{2}{3} x\left(1+x^{2}\right)^{\frac{3}{2}}+\mathrm{C}$
D. $\frac{x^{2}}{2} \sqrt{1+x^{2}}+\frac{1}{2} x^{2} \log \left|x+\sqrt{1+x^{2}}\right|+\mathrm{C}$
Solution:
It is known that, $\int \sqrt{a^{2}+x^{2}} d x=\frac{x}{2} \sqrt{a^{2}+x^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+\mathrm{C}$
$\therefore \int \sqrt{1+x^{2}} d x=\frac{x}{2} \sqrt{1+x^{2}}+\frac{1}{2} \log \left|x+\sqrt{1+x^{2}}\right|+\mathrm{C}$
Hence, the correct answer is A.