Question

Which of the following is(are) correct for $\int \frac{dx}{\sqrt{2-6x-9x^2}}=A{\sin }^{-1}\left(f\right(x\left)\right)+C$
(where $A$ is a fixed constant and $C$ is integration constant)

• A. $A=\frac{1}{3}$
• B. $f\left(x\right)=\sqrt{3}x+\frac{1}{\sqrt{3}}$
• C. $f\left(x\right)=3x+\frac{1}{\sqrt{3}}$
• D. $A=\frac{1}{\sqrt{3}}$

Answer 1

Answer: (A), (B)

$f\left(x\right)=\sqrt{3}x+\frac{1}{\sqrt{3}}$

Let $I=\int \frac{dx}{\sqrt{2-6x-9x^2}}$
$=\int \frac{dx}{\sqrt{3-(3x+1{)}^2}}=\int \frac{dx}{\sqrt{(\sqrt{3}{)}^2-(3x+1{)}^2}}\Rightarrow I=\frac{1}{3}{\sin }^{-1}\left(\frac{3x+1}{\sqrt{3}}\right)+C\therefore A=\frac{1}{3};f\left(x\right)=\left(\frac{3x+1}{\sqrt{3}}\right)$