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

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

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

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

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

Solution steps

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

Long division

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

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

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

\int 3 d x = 3 x

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

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

Add a constant to the solution

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

- cos (2 x)

\int_{1}^{4} \frac{x - 2}{x} d x

f (x) = 3 arccsc (x)

\int x^{2} sech^{2} (x^{3}) d x

\frac{d}{d x} (x^{3} - 8)

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