What is the integral of (x^4+x+1)/(x^3-x) ?

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

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integral of (x^4+x+1)/(x^3-x)

\int \frac{x ^{4} + x + 1}{x ^{3} - x} d x

\frac{1}{2} x^{2} + \frac{1}{2} ln x^{2} - 1 + ln ∣ x - 1 ∣ - ln ∣ x ∣ - \frac{1}{2} + C

Solution steps

\int \frac{x ^{4} + x + 1}{x ^{3} - x} d x

Expand

\frac{x ^{4} + x + 1}{x ^{3} - x} : \frac{x ^{4}}{x ^{3} - x} + \frac{x}{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{x ^{4}}{x ^{3} - x} d x + \int \frac{x}{x ^{3} - x} d x + \int \frac{1}{x ^{3} - x} d x

\int \frac{x ^{4}}{x ^{3} - x} d x = \frac{1}{2} (x^{2} - 1 + ln x^{2} - 1)

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

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

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

Simplify

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

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

Add a constant to the solution

= \frac{1}{2} x^{2} + \frac{1}{2} ln x^{2} - 1 + ln ∣ x - 1 ∣ - ln ∣ x ∣ - \frac{1}{2} + C

y^{^{'}} - y = 2 x

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

\frac{1}{( 2 x + 1 ) ^{2}}

\int \frac{1}{x ^{2} 4 x ^{2} - 25} d x

t^{2} (5 t + 2)^{3}

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