Question

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

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

Answer 1

Answer: (A)

$\frac{x}{2}\sqrt{1+x^2}+\frac{1}{2}\log\mid x+\sqrt{1+x^2}\mid +C$

Answer 2

It is known that,$\int \sqrt{a^2+x^2}dx = \frac{x}{2}\sqrt{a^2+x^2}+\frac{a^2}{2}\log\mid x+\sqrt{x^2+a^2}\mid+C$

∴$\int \sqrt{1+x^2}dx=\frac{x}{2}\sqrt{1+x^2}+\frac{1}{2}\log\mid x+\sqrt{1+x^2}\mid+C$

Hence, the correct answer is A.