What is the integral of (10x^2-9x+16)/(x^3+4x) ?

The integral of (10x^2-9x+16)/(x^3+4x) is -9/2 arctan(x/2)+3ln|x^2+4|+4ln|x|+C

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integral of (10x^2-9x+16)/(x^3+4x)

\int \frac{10 x ^{2} - 9 x + 16}{x ^{3} + 4 x} d x

- \frac{9}{2} arctan (\frac{x}{2}) + 3 ln x^{2} + 4 + 4 ln ∣ x ∣ + C

Solution steps

\int \frac{10 x ^{2} - 9 x + 16}{x ^{3} + 4 x} d x

Expand

\frac{10 x ^{2} - 9 x + 16}{x ^{3} + 4 x} : \frac{10 x ^{2}}{x ^{3} + 4 x} - \frac{9 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{10 x ^{2}}{x ^{3} + 4 x} d x - \int \frac{9 x}{x ^{3} + 4 x} d x + \int \frac{16}{x ^{3} + 4 x} d x

\int \frac{10 x ^{2}}{x ^{3} + 4 x} d x = 5 ln x^{2} + 4

\int \frac{9 x}{x ^{3} + 4 x} d x = \frac{9}{2} 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)

= 5 ln x^{2} + 4 - \frac{9}{2} arctan (\frac{x}{2}) + 16 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4)

Simplify

5 ln x^{2} + 4 - \frac{9}{2} arctan (\frac{x}{2}) + 16 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4) : - \frac{9}{2} arctan (\frac{x}{2}) + 3 ln x^{2} + 4 + 4 ln ∣ x ∣

= - \frac{9}{2} arctan (\frac{x}{2}) + 3 ln x^{2} + 4 + 4 ln ∣ x ∣

Add a constant to the solution

= - \frac{9}{2} arctan (\frac{x}{2}) + 3 ln x^{2} + 4 + 4 ln ∣ x ∣ + C

- 3 y^{^{'}} + \frac{3 y}{x} = x^{3} y^{4}

\frac{d y}{d t} = 2 y + y^{3}, y (0) = 1

x^{\frac{2}{7}}

4 x - 3

y = \frac{4 x}{1 + x ^{2}}, (4, \frac{16}{17})

What is the integral of (10x^2-9x+16)/(x^3+4x) ?
The integral of (10x^2-9x+16)/(x^3+4x) is -9/2 arctan(x/2)+3ln|x^2+4|+4ln|x|+C