What is the integral of (1-x)/(1+x^2) ?

The integral of (1-x)/(1+x^2) is arctan(x)-1/2 ln|1+x^2|+C

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integral of (1-x)/(1+x^2)

\int \frac{1 - x}{1 + x ^{2}} d x

arctan (x) - \frac{1}{2} ln 1 + x^{2} + C

Solution steps

\int \frac{1 - x}{1 + x ^{2}} d x

Expand

\frac{1 - x}{1 + x ^{2}} : \frac{1}{1 + x ^{2}} - \frac{x}{1 + x ^{2}}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{1 + x ^{2}} d x = arctan (x)

\int \frac{x}{1 + x ^{2}} d x = \frac{1}{2} ln 1 + x^{2}

= arctan (x) - \frac{1}{2} ln 1 + x^{2}

Add a constant to the solution

= arctan (x) - \frac{1}{2} ln 1 + x^{2} + C

\frac{d}{d x} ((x^{2} + 1)^{x + 1})

\frac{d}{d x} (c x + c - c^{2})

\int_{0}^{\infty} x \cdot e^{- x} d x

\frac{1}{2 x + 1}, at a = 3

\frac{d y}{d x} = \frac{9}{x}

What is the integral of (1-x)/(1+x^2) ?
The integral of (1-x)/(1+x^2) is arctan(x)-1/2 ln|1+x^2|+C