# Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.1 Exercises - Page 512: 2

$(a)\ln \sqrt {x^{2}+1}+C$

#### Work Step by Step

$\frac{dy}{dx}=\frac{x}{X^{2}+1}$ $\int dy=\int \frac{x}{x^{2}+1}dx$ Let $u=x^{2}+1$. $\frac{du}{dx}=2x$ $dx=\frac{1}{2x}du$ $y=\int (\frac{x}{u})(\frac{1}{2x})du$ $y=\frac{1}{2}\int \frac{du}{u}$ $y=\frac{1}{2}\ln (x^{2}+1)+C$ $y=\ln (x^{2}+1)^{\frac{1}{2}}+C$ $y=\ln \sqrt {x^{2}+1}+C$ Thus the answer is $(a)\ln \sqrt {x^{2}+1}+C$.

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