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

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

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

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

ln ∣ x - 1 ∣ + arctan (x) + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x - 1} d x = ln ∣ x - 1 ∣

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

= ln ∣ x - 1 ∣ + arctan (x)

Add a constant to the solution

= ln ∣ x - 1 ∣ + arctan (x) + C

y^{^{'}} + \frac{1 - x}{2 x} y = 0

f (x) = 2 x + \frac{1936}{x}

f (x) = 110 x - 16 x^{2}, at x = 4

(x^{2} - x y) d x + x d y = 0

\frac{1}{x + 8}

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