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

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

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

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

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

Solution steps

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

Long division

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

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

\int x d x = \frac{x ^{2}}{2}

\int 1 d x = x

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

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

Add a constant to the solution

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

f (x) = e^{e x}

y^{^{''}} - 4 y^{^{'}} + 3 y = 0, y (0) = - 4, y^{^{'}} (0) = 5

\frac{\partial}{\partial x} (33 - 3 x^{2} - 3 y^{2})

f (x) = \frac{x}{x ^{2} - 1}, at x = 2

(5 x^{6} + 2 x^{3})^{4}

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