Question

integrate -x/1+x² dx​

Answer 1

Step-by-step explanation:

We have,

$\int \frac{ - x}{1 + {x}^{2} } dx$

$= - \frac{1}{2} \int \frac{2x}{1 + {x}^{2} } dx$

$let \: \: 1 + {x}^{2} = y \\ \implies2xdx = dy$

$= - \frac{1}{2} \int \frac{dy}{y}$

$= - \frac{1}{2} ln | y | + c$

$= - \frac{1}{2} ln |1 + {x}^{2} | + c$

$= - ln \sqrt{ 1 + {x}^{2} } + c$