Question

Evaluate the integral:

$\int \frac{(a{x}^2-b)dx}{x\sqrt{c^2{x}^2-(a{x}^2+b{)}^2}}$

• A. ${\cos }^{-1}\left(\frac{ax+\frac{b}{x}}{c}\right)+k$
• B. ${\cos }^{-1}\left(\frac{ax-\frac{b}{x}}{c}\right)+k$
• C. ${\sin }^{-1}\left(\frac{ax+\frac{b}{x}}{c}\right)+k$
• D. ${\sin }^{-1}\left(\frac{ax-\frac{b}{x}}{c}\right)+k$

Answer 1

The correct option is C ${\sin }^{-1}\left(\frac{ax+\frac{b}{x}}{c}\right)+k$

$=\int \frac{x^2(a-b/x^2)dx}{xx\sqrt{c^2{x}^2-(a{x}^2+b{)}^2}}$

$=\int \frac{(a-b/x^2)dx}{\sqrt{c^2-(ax+b/x{)}^2}}$

take, $t=(ax+b/x),dt=(a-\frac{b}{x^2})dx$

So, it is

$\Rightarrow \frac{dt}{\sqrt{c^2-t^2}}$ $[\therefore \int \frac{dx}{\sqrt{x^2-a^2}}={\sin }^{-1}\left(\frac{x}{a}\right)+c]$

$={\sin }^{-1}\left(\frac{t}{c}\right)+k$

$={\sin }^{-1}\left(\frac{ax+b/x}{c}\right)+k$

$={\sin }^{-1}\left(\frac{ax+b/x}{c}\right)+k$

So, option (C) is correct.