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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (ln (x + y))

y^{^{'}} = e^{(y - x)}

(\frac{q ^{4}}{4} + \frac{2 q ^{3}}{3} - \frac{11 q ^{2}}{2} + 12 q)^{^{'}}

4 (1 + x^{2}) \frac{d y}{d x} = 2 x y (y^{4} - 1)

f (x) = \frac{3}{4} x

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