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

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

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

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

2 arctan (x) - ln ∣ x ∣ + C

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

= 2 arctan (x) - ln ∣ x ∣

Add a constant to the solution

= 2 arctan (x) - ln ∣ x ∣ + C

\frac{d}{d x} (7 - 5 x)

y = arcsin (\frac{x}{2}), at x = 1

\frac{d x}{d t} = sin (t) x + e^{- c o s (t)} sin (t)

x \to \infty lim (- (5))

3 y^{^{'}} + x y = 5 x, y (9) = 4

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