Question

The value of $\int \frac{\sqrt{x^2-a^2}}{x}dx$ will be

• A. $\sqrt{(x^2-a^2)}-ata{n}^{-1}\left[\frac{\sqrt{(x^2-a^2)}}{a}\right]$
• B. $\sqrt{(x^2-a^2)}+ata{n}^{-1}\left[\frac{\sqrt{(x^2-a^2)}}{a}\right]$
• C. $\sqrt{(x^2-a^2)}+a^{2}ta{n}^{-1}\left[\sqrt{x^2-a^2}\right]$
• D. $\frac{ta{n}^{-1}x}{a+c}$

Answer 1

The correct option is A $\sqrt{(x^2-a^2)}-ata{n}^{-1}\left[\frac{\sqrt{(x^2-a^2)}}{a}\right]$

Let $\sqrt{(x^2-a^2)}=t\Rightarrow x^2-a^2=t^2\Rightarrow x^2=a^2+t^2\therefore xdx=tdt$
$\therefore \int \frac{\sqrt{(x^2-a^2)}}{x}dx=\int \frac{\sqrt{(x^2-a^2)}}{x^2}dx$
$\Rightarrow I=\int \frac{t}{a^2+t^2}tdt=\int \frac{t^2}{a^2+t^2}dt$
$\Rightarrow I=\int (1-\frac{a^2}{a^2+t^2})dt=t-a^2\frac{1}{a}ta{n}^{-1}\left(\frac{t}{a}\right)$
$\Rightarrow I=\sqrt{(x^2-a^2)}-ata{n}^{-1}\left[\frac{\left\{\sqrt{x^2-a^2}\right\}}{a}\right]$