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

The integral of (x^2+4)/(3x^2+4x^2-4x) is 1/7 x+53/49 ln|7x-4|-ln|x|-4/49+C

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

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

\frac{1}{7} x + \frac{53}{49} ln ∣ 7 x - 4 ∣ - ln ∣ x ∣ - \frac{4}{49} + C

Solution steps

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

Expand

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

\int \frac{x ^{2}}{3 x ^{2} + 4 x ^{2} - 4 x} d x = \frac{1}{49} (7 x - 4 + 4 ln ∣ 7 x - 4 ∣)

\int \frac{4}{3 x ^{2} + 4 x ^{2} - 4 x} d x = - ln ∣ x ∣ + ln ∣ 7 x - 4 ∣

= \frac{1}{49} (7 x - 4 + 4 ln ∣ 7 x - 4 ∣) - ln ∣ x ∣ + ln ∣ 7 x - 4 ∣

Simplify

\frac{1}{49} (7 x - 4 + 4 ln ∣ 7 x - 4 ∣) - ln ∣ x ∣ + ln ∣ 7 x - 4 ∣ : \frac{1}{7} x + \frac{53}{49} ln ∣ 7 x - 4 ∣ - ln ∣ x ∣ - \frac{4}{49}

= \frac{1}{7} x + \frac{53}{49} ln ∣ 7 x - 4 ∣ - ln ∣ x ∣ - \frac{4}{49}

Add a constant to the solution

= \frac{1}{7} x + \frac{53}{49} ln ∣ 7 x - 4 ∣ - ln ∣ x ∣ - \frac{4}{49} + C

What is the integral of (x^2+4)/(3x^2+4x^2-4x) ?
The integral of (x^2+4)/(3x^2+4x^2-4x) is 1/7 x+53/49 ln|7x-4|-ln|x|-4/49+C