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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

Add a constant to the solution

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

\frac{d y}{d x} = \frac{( 2 x ( y - 1 ) )}{( x ^{2} + 3 )}

y = 2 x^{3} - x^{2} + 2 (1.3)

(10.4) (4.15)

\int \frac{e ^{x}}{1 + e ^{x}} d x

\int x^{2} + x + \frac{2}{x - 1} d x

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