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

The integral of (x^2+6x)/(x^3-x^2+x-1) is 7/2 ln|x-1|-5/4 ln|x^2+1|+7/2 arctan(x)+C

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

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

\frac{7}{2} ln ∣ x - 1 ∣ - \frac{5}{4} ln x^{2} + 1 + \frac{7}{2} arctan (x) + C

Solution steps

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

Expand

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

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

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

\int \frac{6 x}{x ^{3} - x ^{2} + x - 1} d x = 6 (\frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} (- \frac{1}{2} ln x^{2} + 1 + arctan (x)))

= \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} (\frac{1}{2} ln x^{2} + 1 + arctan (x)) + 6 (\frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} (- \frac{1}{2} ln x^{2} + 1 + arctan (x)))

Simplify

\frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} (\frac{1}{2} ln x^{2} + 1 + arctan (x)) + 6 (\frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} (- \frac{1}{2} ln x^{2} + 1 + arctan (x))) : \frac{7}{2} ln ∣ x - 1 ∣ - \frac{5}{4} ln x^{2} + 1 + \frac{7}{2} arctan (x)

= \frac{7}{2} ln ∣ x - 1 ∣ - \frac{5}{4} ln x^{2} + 1 + \frac{7}{2} arctan (x)

Add a constant to the solution

= \frac{7}{2} ln ∣ x - 1 ∣ - \frac{5}{4} ln x^{2} + 1 + \frac{7}{2} arctan (x) + C

x \to 2 lim ((x + 1) \cdot (x + 3))

\frac{\partial}{\partial x} (x e^{- x} (a + b x + c x^{2}))

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

x \to 2 lim (\frac{1}{x ^{2} - 4})

(x^{2} - 3) (x^{3} - 2 x)

What is the integral of (x^2+6x)/(x^3-x^2+x-1) ?
The integral of (x^2+6x)/(x^3-x^2+x-1) is 7/2 ln|x-1|-5/4 ln|x^2+1|+7/2 arctan(x)+C