What is the integral of (x^4+x-9)/(x^2+3) ?

The integral of (x^4+x-9)/(x^2+3) is (x^3)/3-3x+1/2 ln|x^2+3|+C

integral of (x^4+x-9)/(x^2+3)

\int \frac{x ^{4} + x - 9}{x ^{2} + 3} d x

\frac{x ^{3}}{3} - 3 x + \frac{1}{2} ln x^{2} + 3 + C

Solution steps

\int \frac{x ^{4} + x - 9}{x ^{2} + 3} d x

Long division

\frac{x ^{4} + x - 9}{x ^{2} + 3} : x^{2} - 3 + \frac{x}{x ^{2} + 3}

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

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

\int 3 d x = 3 x

\int \frac{x}{x ^{2} + 3} d x = \frac{1}{2} ln x^{2} + 3

= \frac{x ^{3}}{3} - 3 x + \frac{1}{2} ln x^{2} + 3

Add a constant to the solution

= \frac{x ^{3}}{3} - 3 x + \frac{1}{2} ln x^{2} + 3 + C

\frac{\partial}{\partial x} (x^{3} + y^{3} + z^{3} + 6 x yz)

\int \frac{3 - 4 x ^{3} + 6 x ^{6}}{x ^{6}} d x

- y^{^{'''}} + 2 y^{^{''}} + 3 y^{^{'}} + 3 = 0

\frac{d}{d x} (ln (x + 2 y))

\frac{d}{d x} (cot (11 x))

What is the integral of (x^4+x-9)/(x^2+3) ?
The integral of (x^4+x-9)/(x^2+3) is (x^3)/3-3x+1/2 ln|x^2+3|+C