What is the integral of ((1-u^2-u-u^3))/(1+u^2) ?

The integral of ((1-u^2-u-u^3))/(1+u^2) is -1/2 u^2-u+2arctan(u)-1/2+C

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integral of ((1-u^2-u-u^3))/(1+u^2)

\int \frac{( 1 - u ^{2} - u - u ^{3} )}{1 + u ^{2}} d u

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

Solution steps

\int \frac{1 - u ^{2} - u - u ^{3}}{1 + u ^{2}} d u

Expand

\frac{1 - u ^{2} - u - u ^{3}}{1 + u ^{2}} : \frac{1}{1 + u ^{2}} - \frac{u ^{2}}{1 + u ^{2}} - \frac{u}{1 + u ^{2}} - \frac{u ^{3}}{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 ^{2}}{1 + u ^{2}} d u - \int \frac{u}{1 + u ^{2}} d u - \int \frac{u ^{3}}{1 + u ^{2}} d u

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

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

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

\int \frac{u ^{3}}{1 + u ^{2}} d u = \frac{1}{2} (1 + u^{2} - ln 1 + u^{2})

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

Simplify

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

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

Add a constant to the solution

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

x^{2} - 3

\frac{d}{d z} (ln (\frac{b ^{2} - z ^{2}}{b ^{2} + z ^{2}}))

y^{^{''}} + 9 y = 4 csc^{2} (3 t)

x \to 0 lim (\frac{2 x}{tan ( x )})

\int_{10}^{20} 400 - x^{2} d x

What is the integral of ((1-u^2-u-u^3))/(1+u^2) ?
The integral of ((1-u^2-u-u^3))/(1+u^2) is -1/2 u^2-u+2arctan(u)-1/2+C