# Integration of : int(-1/2x+1/2)/(x^2 +1)dx ?

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

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Feb 14, 2018

#### Answer:

$\int \frac{- \frac{1}{2} x + \frac{1}{2}}{{x}^{2} + 1} \mathrm{dx} = \frac{1}{2} \arctan x - \frac{1}{4} \ln \left(1 + {x}^{2}\right) + C$

#### Explanation:

Simplify the integrand:

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

using the linearity of the integral:

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

solve the two integrals:

$\int \frac{\mathrm{dx}}{1 + {x}^{2}} = \arctan x + {c}_{1}$

$\int \frac{x \mathrm{dx}}{1 + {x}^{2}} = \frac{1}{2} \int \frac{d \left(1 + {x}^{2}\right)}{1 + {x}^{2}} = \frac{1}{2} \ln \left(1 + {x}^{2}\right) + {c}_{2}$

and substituting the partial solutions:

$\int \frac{- \frac{1}{2} x + \frac{1}{2}}{{x}^{2} + 1} \mathrm{dx} = \frac{1}{2} \arctan x - \frac{1}{4} \ln \left(1 + {x}^{2}\right) + C$

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