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

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

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

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

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

Solution steps

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

Expand

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

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

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

\int \frac{4}{x ^{3} + 4 x} d x = ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 4

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

Simplify

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

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

Add a constant to the solution

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

y^{^{''}} + 2 y^{^{'}} + y = 2 e^{- x}

\frac{d}{d x} (36 x^{2} - 72 x)

x \to 0 lim (cos (\frac{π}{x}))

\int \frac{ln ( x )}{x ^{14}} d x

\int (1 + sin (θ))^{2} d θ

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