What is the integral of 8x^2+6+9/(x^2+1) ?

The integral of 8x^2+6+9/(x^2+1) is (8x^3)/3+6x+9arctan(x)+C

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integral of 8x^2+6+9/(x^2+1)

\int 8 x^{2} + 6 + \frac{9}{x ^{2} + 1} d x

\frac{8 x ^{3}}{3} + 6 x + 9 arctan (x) + C

Solution steps

\int 8 x^{2} + 6 + \frac{9}{x ^{2} + 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

= \int 8 x^{2} d x + \int 6 d x + \int \frac{9}{x ^{2} + 1} d x

\int 8 x^{2} d x = \frac{8 x ^{3}}{3}

\int 6 d x = 6 x

\int \frac{9}{x ^{2} + 1} d x = 9 arctan (x)

= \frac{8 x ^{3}}{3} + 6 x + 9 arctan (x)

Add a constant to the solution

= \frac{8 x ^{3}}{3} + 6 x + 9 arctan (x) + C

y^{^{''}} + 16 y^{^{'}} + 73 y = 0

y^{^{''}} - 8 y^{^{'}} + 4 y = 0

\frac{d y}{d x} = 2

\int e^{- x^{8}} (- 8 x^{7}) d x

x \to 16 lim (\frac{x ^{2}}{2} - \frac{8}{x})

What is the integral of 8x^2+6+9/(x^2+1) ?
The integral of 8x^2+6+9/(x^2+1) is (8x^3)/3+6x+9arctan(x)+C