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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d r}{d θ} + r tan (θ) = 3 tan (θ)

e^{5} + ln (7)

\int \frac{1}{3 x + 4} d x

y^{^{'}} + x y = a

\frac{d}{d x} (- 3 x^{5})

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