What is the integral of (8x^2-7x+16)/(x^3+4x) ?

The integral of (8x^2-7x+16)/(x^3+4x) is -7/2 arctan(x/2)+2ln|x^2+4|+4ln|x|+C

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integral of (8x^2-7x+16)/(x^3+4x)

\int \frac{8 x ^{2} - 7 x + 16}{x ^{3} + 4 x} d x

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

Solution steps

\int \frac{8 x ^{2} - 7 x + 16}{x ^{3} + 4 x} d x

Expand

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

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

Simplify

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

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

Add a constant to the solution

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

\int_{1}^{3} \frac{1}{x ^{3}} d x

x 3 - x^{2}

x \to 2 lim (\frac{1}{∣ 2 - x ∣})

y = - x^{2} + 8 x - 12, y = x^{2} - 8 x + 18, x = 3 x = 6

f (x) = 6 x^{2} + 4 x - 7, - 3

What is the integral of (8x^2-7x+16)/(x^3+4x) ?
The integral of (8x^2-7x+16)/(x^3+4x) is -7/2 arctan(x/2)+2ln|x^2+4|+4ln|x|+C