What is the integral of (6x^2+x-1)/(3x^3-9x^2+x-3) ?

The integral of (6x^2+x-1)/(3x^3-9x^2+x-3) is 2ln|x-3|+1/(sqrt(3))arctan(sqrt(3)x)+C

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integral of (6x^2+x-1)/(3x^3-9x^2+x-3)

\int \frac{6 x ^{2} + x - 1}{3 x ^{3} - 9 x ^{2} + x - 3} d x

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

Solution steps

\int \frac{6 x ^{2} + x - 1}{3 x ^{3} - 9 x ^{2} + x - 3} d x

Take the partial fraction of

\frac{6 x ^{2} + x - 1}{3 x ^{3} - 9 x ^{2} + x - 3} : \frac{2}{x - 3} + \frac{1}{3 x ^{2} + 1}

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

\int \frac{2}{x - 3} d x = 2 ln ∣ x - 3 ∣

\int \frac{1}{3 x ^{2} + 1} d x = \frac{1}{3} arctan (3 x)

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

Add a constant to the solution

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

tan (e^{6 t}) + e^{t a n (6 t)}

\int - e^{2 t} + t e^{t} d t

\int_{0}^{1} x 6 - x^{2} d x

\frac{d}{d x} (sin (x) + \frac{3}{4} cot (x))

\frac{\partial}{\partial x} (cos (x y^{2}))

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