Mathematics

Integrate:
$$\displaystyle \int \dfrac{\sqrt{1+x^{2}}}{x}dx$$


SOLUTION
$$\int \dfrac{\sqrt{1+x^2}}{x}dx$$

substitute  $$u=\sqrt{1+x^2}\rightarrow dx=\dfrac{du}{u^2-1}$$

$$=\int \dfrac{u^2}{u^2-1}du$$

$$=-\int \dfrac{u^2}{1-u^2}du$$

$$=-\int  \left ( \dfrac{1}{1-u^2}-1 \right )du$$

$$=-\left ( \dfrac{\ln |u+1|}{2} -\dfrac{\ln |u-1|}{2}-u\right )$$

$$=-\left ( \dfrac{\ln |\sqrt{1+x^2}+1|}{2} -\dfrac{\ln |\sqrt{1+x^2}-1|}{2}-\sqrt{1+x^2}\right )+C$$
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