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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.