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

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Indefinite Integration

Question:

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

Options:

$\frac{x}{2}\sqrt{1+x^{2}}+\frac{1}{2}\log \left|\left(x+\sqrt{1+x^{2}}\right)\right|+C$

$\frac{2}{3}\left(1+x^{2}\right)^{\frac{3}{2}}+C$

$\frac{2}{3}x\left(1+x{2}\right)^{\frac{3}{2}}+C$

$\frac{x^{2}}{2}\sqrt{1+x^{2}}+\frac{1}{2}x^{2}\log \left|x+\sqrt{1+x^{2}}\right|+C$

Correct Answer:

$\frac{x}{2}\sqrt{1+x^{2}}+\frac{1}{2}\log \left|\left(x+\sqrt{1+x^{2}}\right)\right|+C$

Explanation:

$\int \sqrt{x^{2}+a^{2}}dx=\frac{1}{2}x\sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2}\log \left|x+\sqrt{x^{2}+a^{2}}\right|+C$