What is the integral of (7x^2-6x+16)/(x^3+4x) ?

The integral of (7x^2-6x+16)/(x^3+4x) is 3/2 ln|x^2+4|-3arctan(x/2)+4ln|x|+C

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integral of (7x^2-6x+16)/(x^3+4x)

\int \frac{7 x ^{2} - 6 x + 16}{x ^{3} + 4 x} d x

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

Solution steps

\int \frac{7 x ^{2} - 6 x + 16}{x ^{3} + 4 x} d x

Expand

\frac{7 x ^{2} - 6 x + 16}{x ^{3} + 4 x} : \frac{7 x ^{2}}{x ^{3} + 4 x} - \frac{6 x}{x ^{3} + 4 x} + \frac{16}{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{16}{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{16}{x ^{3} + 4 x} d x = 16 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4)

= \frac{7}{2} ln x^{2} + 4 - 3 arctan (\frac{x}{2}) + 16 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4)

Simplify

\frac{7}{2} ln x^{2} + 4 - 3 arctan (\frac{x}{2}) + 16 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4) : \frac{3}{2} ln x^{2} + 4 - 3 arctan (\frac{x}{2}) + 4 ln ∣ x ∣

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

Add a constant to the solution

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

2 y^{^{''}} + 8 y^{^{'}} + 80 y = 80 cos (8 t)

x \to - 3 lim (x + 3)

(x - 8) y^{^{'}} + y = x^{2} - 6

z \to 0 lim (\frac{z}{tan ( z )})

\frac{d}{d t} (2 t + 7)

What is the integral of (7x^2-6x+16)/(x^3+4x) ?
The integral of (7x^2-6x+16)/(x^3+4x) is 3/2 ln|x^2+4|-3arctan(x/2)+4ln|x|+C