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{\partial}{\partial x} (4 x (x^{2} + y^{2}) - 4 a^{2} x)

\frac{a}{x}

\frac{s}{( s - 1 )}

y (1 + x^{2}) y^{^{'}} - x (6 + y^{2}) = 0

y = 5 x^{4} e^{x} + e^{x} x^{5}

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