Question

I.F of $(1+y^2)dx=(ta{n}^{-1}y-x)dy$ is:

• A. $e^{ta{n}^{-1}x}$
• B. $ta{n}^{-1}x$
• C. $ta{n}^{-1}y$
• D. $e^{ta{n}^{-1}y}$

Answer 1

The correct option is D $e^{ta{n}^{-1}y}$

$\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{{\tan }^{-1}y}{1+y^2}$
$\Rightarrow I.F=e^{\int P\left(y\right)dy}=e^{\int \frac{dy}{1+y^2}}=e^{{\tan }^{-1}y}$