What is the integral of (6x^4+24x^2+4)/(x^3+4x) ?

The integral of (6x^4+24x^2+4)/(x^3+4x) is 3x^2-1/2 ln|x^2+4|+ln|x|+12+C

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integral of (6x^4+24x^2+4)/(x^3+4x)

\int \frac{6 x ^{4} + 24 x ^{2} + 4}{x ^{3} + 4 x} d x

3 x^{2} - \frac{1}{2} ln x^{2} + 4 + ln ∣ x ∣ + 12 + C

Solution steps

\int \frac{6 x ^{4} + 24 x ^{2} + 4}{x ^{3} + 4 x} d x

Expand

\frac{6 x ^{4} + 24 x ^{2} + 4}{x ^{3} + 4 x} : \frac{6 x ^{4}}{x ^{3} + 4 x} + \frac{24 x ^{2}}{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{6 x ^{4}}{x ^{3} + 4 x} d x + \int \frac{24 x ^{2}}{x ^{3} + 4 x} d x + \int \frac{4}{x ^{3} + 4 x} d x

\int \frac{6 x ^{4}}{x ^{3} + 4 x} d x = 3 (x^{2} + 4 - 4 ln x^{2} + 4)

\int \frac{24 x ^{2}}{x ^{3} + 4 x} d x = 12 ln x^{2} + 4

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

= 3 (x^{2} + 4 - 4 ln x^{2} + 4) + 12 ln x^{2} + 4 + ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 4

Simplify

3 (x^{2} + 4 - 4 ln x^{2} + 4) + 12 ln x^{2} + 4 + ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 4 : 3 x^{2} - \frac{1}{2} ln x^{2} + 4 + ln ∣ x ∣ + 12

= 3 x^{2} - \frac{1}{2} ln x^{2} + 4 + ln ∣ x ∣ + 12

Add a constant to the solution

= 3 x^{2} - \frac{1}{2} ln x^{2} + 4 + ln ∣ x ∣ + 12 + C

f (x) = 2 x + 3

\frac{( s + 2 )}{( s ^{2} + 9 )}

\frac{d}{d x} (\frac{( x + 1 ) ^{2}}{x})

y = x^{3}, y = 2 x^{2}

What is the integral of (6x^4+24x^2+4)/(x^3+4x) ?
The integral of (6x^4+24x^2+4)/(x^3+4x) is 3x^2-1/2 ln|x^2+4|+ln|x|+12+C