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

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

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integral of 1-(2x)/pi

\int 1 - \frac{2 x}{π} d x

x - \frac{x ^{2}}{π} + C

Solution steps

\int 1 - \frac{2 x}{π} 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 1 d x - \int \frac{2 x}{π} d x

\int 1 d x = x

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

= x - \frac{x ^{2}}{π}

Add a constant to the solution

= x - \frac{x ^{2}}{π} + C