What is the integral of 2/(x^3-5x^2+3x+9) ?

The integral of 2/(x^3-5x^2+3x+9) is 2(1/16 ln|x+1|-1/16 ln|x-3|-1/(4(x-3)))+C

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integral of 2/(x^3-5x^2+3x+9)

\int \frac{2}{x ^{3} - 5 x ^{2} + 3 x + 9} d x

2 (\frac{1}{16} ln ∣ x + 1 ∣ - \frac{1}{16} ln ∣ x - 3 ∣ - \frac{1}{4 ( x - 3 )}) + C

Solution steps

\int \frac{2}{x ^{3} - 5 x ^{2} + 3 x + 9} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{x ^{3} - 5 x ^{2} + 3 x + 9} d x

Take the partial fraction of

\frac{1}{x ^{3} - 5 x ^{2} + 3 x + 9} : \frac{1}{16 ( x + 1 )} - \frac{1}{16 ( x - 3 )} + \frac{1}{4 ( x - 3 ) ^{2}}

= 2 \cdot \int \frac{1}{16 ( x + 1 )} - \frac{1}{16 ( x - 3 )} + \frac{1}{4 ( x - 3 ) ^{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

= 2 (\int \frac{1}{16 ( x + 1 )} d x - \int \frac{1}{16 ( x - 3 )} d x + \int \frac{1}{4 ( x - 3 ) ^{2}} d x)

\int \frac{1}{16 ( x + 1 )} d x = \frac{1}{16} ln ∣ x + 1 ∣

\int \frac{1}{16 ( x - 3 )} d x = \frac{1}{16} ln ∣ x - 3 ∣

\int \frac{1}{4 ( x - 3 ) ^{2}} d x = - \frac{1}{4 ( x - 3 )}

= 2 (\frac{1}{16} ln ∣ x + 1 ∣ - \frac{1}{16} ln ∣ x - 3 ∣ - \frac{1}{4 ( x - 3 )})

Add a constant to the solution

= 2 (\frac{1}{16} ln ∣ x + 1 ∣ - \frac{1}{16} ln ∣ x - 3 ∣ - \frac{1}{4 ( x - 3 )}) + C

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

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

\int + \frac{4 - 32 x}{1 - 16 x ^{2}} d x

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

x^{2} \frac{d y}{d x} + x y = 1

What is the integral of 2/(x^3-5x^2+3x+9) ?
The integral of 2/(x^3-5x^2+3x+9) is 2(1/16 ln|x+1|-1/16 ln|x-3|-1/(4(x-3)))+C