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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

y (θ) = \frac{π}{9} cos (θ)

x \to 0 lim (\frac{1}{x ^{2} + x})

\int - \frac{8}{y ^{9}} d y

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

y = x^{2} - 1, y = - x + 2, x = 0, x = 1

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