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I would be inclined to do this like Random Variable but since you got this far, sec(u)= 1/cos(u) and tan(u)= sin(u)/cos(u) so that [tex]\frac{sec^3(u)}{tan(u)}= \frac{1}{cos^3(u)}\frac{cos(u)}{sin(u)}= \frac{1}{sin(u)cos^2(u)}[/tex]which has sine to an odd power. We can multiply both numerator and denominator by sin(x) to get [tex]\frac{sin(x)}{sin^2(x)cos^2(x)}= \frac{sin(x)}{(1- cos^2(x))cos^2(x)}[/tex]how do u integrate sqrt(1 + x^2) / x??
i reduced this to sec^3(u)/tan(u) but how do i go from here??
[tex]\int \frac{\sqrt{1 + x^2}}{x}\,dx[/tex]
[tex]\text{Let }\,x \,=\,\tan\theta \quad\Rightarrow\quad dx \,=\,\sec^2\!\theta\,d\theta[/tex]
[tex]\text{I reduced this to: }\:\int \frac{\sec^3\!\theta}{\tan\theta}\,d\theta[/tex] . Good!