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

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

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

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

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

Solution steps

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

Expand

\frac{1 - u}{1 + u ^{2}} : \frac{1}{1 + u ^{2}} - \frac{u}{1 + u ^{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 + u ^{2}} d u - \int \frac{u}{1 + u ^{2}} d u

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

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

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

Add a constant to the solution

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

\int x^{2} (x^{3} + 5)^{8} cos ((x^{3} + 5)^{9}) d x

\frac{\partial}{\partial x} (y + x^{2})

x y^{^{'}} + 2 y = - 4 x

\frac{d}{d x} ((x^{3})^{5})

\int \frac{1}{u 2 - u ^{2}} d u

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