What is the integral of (7x^2+3x+4)/(x^3-x) ?

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

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integral of (7x^2+3x+4)/(x^3-x)

\int \frac{7 x ^{2} + 3 x + 4}{x ^{3} - x} d x

\frac{7}{2} ln x^{2} - 1 + \frac{1}{2} ln ∣ x + 1 ∣ + \frac{7}{2} ln ∣ x - 1 ∣ - 4 ln ∣ x ∣ + C

Solution steps

\int \frac{7 x ^{2} + 3 x + 4}{x ^{3} - x} d x

Expand

\frac{7 x ^{2} + 3 x + 4}{x ^{3} - x} : \frac{7 x ^{2}}{x ^{3} - x} + \frac{3 x}{x ^{3} - x} + \frac{4}{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{7 x ^{2}}{x ^{3} - x} d x + \int \frac{3 x}{x ^{3} - x} d x + \int \frac{4}{x ^{3} - x} d x

\int \frac{7 x ^{2}}{x ^{3} - x} d x = \frac{7}{2} ln x^{2} - 1

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

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

= \frac{7}{2} ln x^{2} - 1 - 3 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣) + 4 (- ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣)

Simplify

\frac{7}{2} ln x^{2} - 1 - 3 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣) + 4 (- ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣) : \frac{7}{2} ln x^{2} - 1 + \frac{1}{2} ln ∣ x + 1 ∣ + \frac{7}{2} ln ∣ x - 1 ∣ - 4 ln ∣ x ∣

= \frac{7}{2} ln x^{2} - 1 + \frac{1}{2} ln ∣ x + 1 ∣ + \frac{7}{2} ln ∣ x - 1 ∣ - 4 ln ∣ x ∣

Add a constant to the solution

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

\frac{2}{3 x + 1}, 2

\frac{\partial}{\partial x} (cos (y) e^{x y} + e^{x y} sin (y) x)

x \to 0 lim (\frac{4}{x - 1})

\int \frac{x e ^{x^{2}}}{1 + e ^{x^{2}}} d x

f (x) = 2 x^{2} - 4 x

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