What is the integral of-xarctan(x) ?

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

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integral of-xarctan(x)

\int - x arctan (x) d x

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

Solution steps

\int - x arctan (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int x arctan (x) d x

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int_{0}^{1} x^{2} + 1 d x

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

\frac{\partial}{\partial x} (2 y e^{4 x})

\int (80 - 3 t - (2 t + 30)) d t

\frac{d}{d x} (\frac{x - x ^{2}}{2 - 3 x + x ^{2}})

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