Question

$\int \sqrt{1+x^2}dx$ is equal to

• A. $\frac{x}{2}\sqrt{1+x^2}+\frac{1}{2}log|+\sqrt{1+x^2}|+C$
• B. $\frac{2}{3}(1+x^2{)}^{\frac{2}{3}}+C$
• C. $\frac{2}{3}x(1+x^2{)}^{\frac{3}{2}}+C$
• D. $\frac{x^2}{2}\sqrt{1+x^2}+\frac{1}{2}{x}^{2}log|x+\sqrt{1+x^2}|+C$

Answer 1

The correct option is A $\frac{x}{2}\sqrt{1+x^2}+\frac{1}{2}log|+\sqrt{1+x^2}|+C$

We know that, $\int \sqrt{a^2+x^2}dx=\frac{x}{2}\sqrt{a^2+x^2}+\frac{a^2}{2}\log |x+\sqrt{x^2+a^2}|+C$
$\therefore \int \sqrt{1+x^2}dx=\frac{x}{2}\sqrt{1+x^2}+\frac{1}{2}\log |x+\sqrt{1+x^2}|+C$
Hence, the correct Answer is A.