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

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

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

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

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

Solution steps

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

Expand

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

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

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

\int \frac{6}{x ^{3} + 4 x} d x = 6 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4)

= ln x^{2} + 4 + \frac{3}{2} arctan (\frac{x}{2}) + 6 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4)

Simplify

ln x^{2} + 4 + \frac{3}{2} arctan (\frac{x}{2}) + 6 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4) : \frac{3}{2} arctan (\frac{x}{2}) + \frac{3}{2} ln ∣ x ∣ + \frac{1}{4} ln x^{2} + 4

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

Add a constant to the solution

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

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

y = ln (x + 6)

\int (sin (x))^{4} d x

x^{(\frac{1}{3})}, 8

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