Question

Calcule: ∫$\frac{\sqrt{ 1+x^2}}{x}$ dx

Answer 1

Temos que

$\displaystyle\int\dfrac{\sqrt{1+x^2}}{x}dx, \ \ x=tg (\theta) \Rightarrow dx =sec^2(\theta) d\theta \\ \\ \Rightarrow \sqrt{1+x^2}=\sqrt{1+tg^2(\theta)}=sec(\theta)$

Logo,

$\displaystyle\int\dfrac{\sqrt{1+x^2}}{x}dx=\displaystyle\int\dfrac{sec(\theta)}{tg(\theta)}sec^2(\theta)d\theta=\displaystyle\int\dfrac{sec(\theta)}{tg(\theta)}(tg^2(\theta)+1)d\theta \\  \\ =\displaystyle\int sec(\theta)\begin{pmatrix}\dfrac{tg^2(\theta)}{tg(\theta)}+\dfrac{1}{tg(\theta)}\end{pmatrix}d\theta=\displaystyle\int\begin{pmatrix}sec(\theta)tg(\theta)+\dfrac{sec(\theta)}{tg(\theta)}\end{pmatrix}d\theta \\ \\ =\displaystyle\int (sec(\theta)tg(\theta)+csc(\theta))d\theta$

Logo,

$\displaystyle\int\dfrac{\sqrt{1+x^2}}{x}dx=sec(\theta)+ln|csc(\theta)-cotg(\theta)|$

Mas, sabemos que

$x=tg(\theta)$

Com essa relação e usando a trigonometria no triângulo retângulo, temos que nesse triângulo os catetos valem 1 e x, logo a hipotenusa vale raiz(1+x^2). Portanto, temos que:

$\displaystyle\int\dfrac{\sqrt{1+x^2}}{x}dx=sec(\theta)+ln|csc(\theta)-cotg(\theta)| \\ =\sqrt{1+x^2}+ln\begin{vmatrix}\dfrac{\sqrt{1+x^2}}{x}-\dfrac{1}{x}\end{vmatrix}+C$