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

f (x) = 2 x + 45, at x = 2

49 + 23 sin (\frac{2 π}{52} (t - 10))

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

y^{^{'}} = \frac{2 x}{y}

\int 4 x - 6 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