What is the integral of (20(1-x^2))/(x^2+1) ?

The integral of (20(1-x^2))/(x^2+1) is 20(2arctan(x)-x)+C

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integral of (20(1-x^2))/(x^2+1)

\int \frac{20 ( 1 - x ^{2} )}{x ^{2} + 1} d x

20 (2 arctan (x) - x) + C

Solution steps

\int \frac{20 ( 1 - x ^{2} )}{x ^{2} + 1} d x

Take the constant out:

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

= 20 \cdot \int \frac{1 - x ^{2}}{x ^{2} + 1} d x

\frac{1 - x ^{2}}{x ^{2} + 1} = \frac{2}{x ^{2} + 1} - 1

= 20 \cdot \int \frac{2}{x ^{2} + 1} - 1 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

= 20 (\int \frac{2}{x ^{2} + 1} d x - \int 1 d x)

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

\int 1 d x = x

= 20 (2 arctan (x) - x)

Add a constant to the solution

= 20 (2 arctan (x) - x) + C

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What is the integral of (20(1-x^2))/(x^2+1) ?
The integral of (20(1-x^2))/(x^2+1) is 20(2arctan(x)-x)+C