What is the integral of pi/4 x-xarctan(x) ?

The integral of pi/4 x-xarctan(x) is (pix^2)/8-1/2 x^2arctan(x)-1/2 arctan(x)+1/2 x+C

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integral of pi/4 x-xarctan(x)

\int \frac{π}{4} x - x arctan (x) d x

\frac{π x ^{2}}{8} - \frac{1}{2} x^{2} arctan (x) - \frac{1}{2} arctan (x) + \frac{1}{2} x + C

Solution steps

\int \frac{π}{4} x - x arctan (x) d x

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{π}{4} x d x - \int x arctan (x) d x

\int \frac{π}{4} x d x = \frac{π x ^{2}}{8}

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

= \frac{π x ^{2}}{8} - (\frac{1}{2} x^{2} arctan (x) - \frac{1}{2} (- arctan (x) + x))

Simplify

\frac{π x ^{2}}{8} - (\frac{1}{2} x^{2} arctan (x) - \frac{1}{2} (- arctan (x) + x)) : \frac{π x ^{2}}{8} - \frac{1}{2} x^{2} arctan (x) - \frac{1}{2} arctan (x) + \frac{1}{2} x

= \frac{π x ^{2}}{8} - \frac{1}{2} x^{2} arctan (x) - \frac{1}{2} arctan (x) + \frac{1}{2} x

Add a constant to the solution

= \frac{π x ^{2}}{8} - \frac{1}{2} x^{2} arctan (x) - \frac{1}{2} arctan (x) + \frac{1}{2} x + C

What is the integral of pi/4 x-xarctan(x) ?
The integral of pi/4 x-xarctan(x) is (pix^2)/8-1/2 x^2arctan(x)-1/2 arctan(x)+1/2 x+C