What is the integral of (7x^2-6x+8)/(x^3+4x) ?

The integral of (7x^2-6x+8)/(x^3+4x) is 5/2 ln|x^2+4|-3arctan(x/2)+2ln|x|+C

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integral of (7x^2-6x+8)/(x^3+4x)

\int \frac{7 x ^{2} - 6 x + 8}{x ^{3} + 4 x} d x

\frac{5}{2} ln x^{2} + 4 - 3 arctan (\frac{x}{2}) + 2 ln ∣ x ∣ + C

Solution steps

\int \frac{7 x ^{2} - 6 x + 8}{x ^{3} + 4 x} d x

Expand

\frac{7 x ^{2} - 6 x + 8}{x ^{3} + 4 x} : \frac{7 x ^{2}}{x ^{3} + 4 x} - \frac{6 x}{x ^{3} + 4 x} + \frac{8}{x ^{3} + 4 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} + 4 x} d x - \int \frac{6 x}{x ^{3} + 4 x} d x + \int \frac{8}{x ^{3} + 4 x} d x

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

\int \frac{6 x}{x ^{3} + 4 x} d x = 3 arctan (\frac{x}{2})

\int \frac{8}{x ^{3} + 4 x} d x = 2 ln ∣ x ∣ - ln x^{2} + 4

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

Add similar elements:

\frac{7}{2} ln x^{2} + 4 - ln x^{2} + 4 = \frac{5}{2} ln x^{2} + 4

= \frac{5}{2} ln x^{2} + 4 - 3 arctan (\frac{x}{2}) + 2 ln ∣ x ∣

Add a constant to the solution

= \frac{5}{2} ln x^{2} + 4 - 3 arctan (\frac{x}{2}) + 2 ln ∣ x ∣ + C

\frac{d}{d x} (x sin (x y))

t \to \infty lim (cos (t))

\frac{d y}{d t} = \frac{3 y}{t} + t^{5}

x \to c lim (f (x))

\frac{d}{d x} (- 2 \cdot x \cdot e^{- x^{2}})

What is the integral of (7x^2-6x+8)/(x^3+4x) ?
The integral of (7x^2-6x+8)/(x^3+4x) is 5/2 ln|x^2+4|-3arctan(x/2)+2ln|x|+C