What is the integral of (x^2+4x-5)/(x^3-1) ?

The integral of (x^2+4x-5)/(x^3-1) is 1/2 ln|4x^2+4x+4|+3sqrt(3)arctan(1/(sqrt(3))(2x+1))+C

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integral of (x^2+4x-5)/(x^3-1)

\int \frac{x ^{2} + 4 x - 5}{x ^{3} - 1} d x

\frac{1}{2} ln 4 x^{2} + 4 x + 4 + 33 arctan (\frac{1}{3} (2 x + 1)) + C

Solution steps

\int \frac{x ^{2} + 4 x - 5}{x ^{3} - 1} d x

Simplify

\frac{x ^{2} + 4 x - 5}{x ^{3} - 1} : \frac{x + 5}{x ^{2} + x + 1}

= \int \frac{x + 5}{x ^{2} + x + 1} d x

Expand

\frac{x + 5}{x ^{2} + x + 1} : \frac{x}{x ^{2} + x + 1} + \frac{5}{x ^{2} + x + 1}

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}{x ^{2} + x + 1} d x + \int \frac{5}{x ^{2} + x + 1} d x

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

\int \frac{5}{x ^{2} + x + 1} d x = \frac{10}{3} arctan (\frac{2 x + 1}{3})

= \frac{1}{2} ln 4 x^{2} + 4 x + 4 - \frac{1}{3} arctan (\frac{2 x + 1}{3}) + \frac{10}{3} arctan (\frac{2 x + 1}{3})

Simplify

\frac{1}{2} ln 4 x^{2} + 4 x + 4 - \frac{1}{3} arctan (\frac{2 x + 1}{3}) + \frac{10}{3} arctan (\frac{2 x + 1}{3}) : \frac{1}{2} ln 4 x^{2} + 4 x + 4 + 33 arctan (\frac{1}{3} (2 x + 1))

= \frac{1}{2} ln 4 x^{2} + 4 x + 4 + 33 arctan (\frac{1}{3} (2 x + 1))

Add a constant to the solution

= \frac{1}{2} ln 4 x^{2} + 4 x + 4 + 33 arctan (\frac{1}{3} (2 x + 1)) + C

f (x) = 1680 (2 x + 5)^{4}

x arcsin (x) + 1 - x^{2}

\frac{\partial}{\partial x} (x - ln (x + 1))

\int \frac{7 sin ^{3} ( x )}{x} d x

X^{2}

What is the integral of (x^2+4x-5)/(x^3-1) ?
The integral of (x^2+4x-5)/(x^3-1) is 1/2 ln|4x^2+4x+4|+3sqrt(3)arctan(1/(sqrt(3))(2x+1))+C