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

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

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

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

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

Solution steps

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

Expand

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

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

\int \frac{7 x}{x ^{3} + 4 x} d x = \frac{7}{2} arctan (\frac{x}{2})

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

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

Simplify

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

Add a constant to the solution

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

y^{^{'}} - 2 y = 1

(lo g_{2} (x))^{^{'}}

e^{x^{2}} - 1, 0

x \to - 2 lim (8 - x^{3})

\int \frac{1}{cosh ^{2} ( x )} d x

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