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# Choose the correct answer in $\Large \int \normalsize \sqrt{1+x^2}dx$ is equal to $\begin{array}{l}(A)\;\frac{x}{2}\sqrt{1+x^2}+\frac{1}{2}log\LARGE\mid\normalsize\bigg(x+\sqrt{1+x^2}\bigg)\LARGE\mid\normalsize+C \\(B)\;\frac{2}{3}(1+x^2)^\frac{3}{2}+C\qquad \\ (C)\;\frac{2}{3}x(1+x^2)^\frac{3}{2}+ C \\ (D)\;\frac{x^2}{2}\sqrt{1+x^2}+\frac{1}{2}x^2log\LARGE\mid\normalsize x+\sqrt{1+x^2}\LARGE\mid\normalsize+C \end{array}$

Toolbox:
• $\int \sqrt{a^2+x^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}log |x+\sqrt{x^2+a^2}$
Given $I=\int \sqrt{1+x^2} dx$

This is of the form $\int \sqrt{a^2+x^2}dx$

$=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}log |x+\sqrt{x^2+a^2}|+c$

Here x=x and a=1

Therefore $\int \sqrt{1+x^2}dx=\frac{x}{2}\sqrt {x^2+1}+\frac{1}{2}log |x+\sqrt{x^2+1}|$

Hence A is the correct answer