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

The integral of ((x^2+4))/(x(x+2)(3x-2)) is -ln|x|+1/2 ln|x+2|+5/6 ln|3x-2|+C

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

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

- ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 2 ∣ + \frac{5}{6} ln ∣ 3 x - 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2} + 4}{x ( x + 2 ) ( 3 x - 2 )} : - \frac{1}{x} + \frac{1}{2 ( x + 2 )} + \frac{5}{2 ( 3 x - 2 )}

= \int - \frac{1}{x} + \frac{1}{2 ( x + 2 )} + \frac{5}{2 ( 3 x - 2 )} d 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{1}{x} d x + \int \frac{1}{2 ( x + 2 )} d x + \int \frac{5}{2 ( 3 x - 2 )} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{1}{2 ( x + 2 )} d x = \frac{1}{2} ln ∣ x + 2 ∣

\int \frac{5}{2 ( 3 x - 2 )} d x = \frac{5}{6} ln ∣ 3 x - 2 ∣

= - ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 2 ∣ + \frac{5}{6} ln ∣ 3 x - 2 ∣

Add a constant to the solution

= - ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 2 ∣ + \frac{5}{6} ln ∣ 3 x - 2 ∣ + C

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What is the integral of ((x^2+4))/(x(x+2)(3x-2)) ?
The integral of ((x^2+4))/(x(x+2)(3x-2)) is -ln|x|+1/2 ln|x+2|+5/6 ln|3x-2|+C