What is the integral of 6/(x^2+1)-x^2 ?

The integral of 6/(x^2+1)-x^2 is 6arctan(x)-(x^3)/3+C

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integral of 6/(x^2+1)-x^2

\int \frac{6}{x ^{2} + 1} - x^{2} d x

6 arctan (x) - \frac{x ^{3}}{3} + C

Solution steps

\int \frac{6}{x ^{2} + 1} - x^{2} 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{6}{x ^{2} + 1} d x - \int x^{2} d x

\int \frac{6}{x ^{2} + 1} d x = 6 arctan (x)

\int x^{2} d x = \frac{x ^{3}}{3}

= 6 arctan (x) - \frac{x ^{3}}{3}

Add a constant to the solution

= 6 arctan (x) - \frac{x ^{3}}{3} + C

\frac{\partial}{\partial x} (ln (1 + \frac{1}{x}))

\frac{\partial}{\partial x} (x^{2} + 2 x y + 3 y^{2})

\int \frac{1}{u} d u

y^{^{''}} - 2 y^{^{'}} - 15 y = 0

(4 y + y x^{2}) d y = (2 x + x y^{2}) d x

What is the integral of 6/(x^2+1)-x^2 ?
The integral of 6/(x^2+1)-x^2 is 6arctan(x)-(x^3)/3+C