Question

Choose the correct answer in the given question.
$\int \sqrt{1+x^2}dx$ is equal to
$\left(a\right)\frac{x}{2}\sqrt{1+x^2}+\frac{1}{2}log|x+\sqrt{1+x^2}|+C\left(b\right)\frac{2}{3}(1+x^2{)}^{\frac{3}{2}}+C(c\left)\frac{2}{3}x\right(1+x^2{)}^{\frac{3}{2}}+C\left(d\right)\frac{x^2}{2}\sqrt{1+x^2}+\frac{1}{2}{x}^{2}log|x+\sqrt{1+x^2}|$

Answer 1

Let $I=\int \sqrt{1+x^2}dx$
$\Rightarrow I=\frac{x}{2}\sqrt{1+x^2}+\frac{1}{2}log|x+\sqrt{1+x^2}|+C[\because \int \sqrt{x^2+a^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}log|x+\sqrt{x^2+a^2}|]$
Hence, the correct option is (a).