What is the integral of (12x^2-4x+7)/(3x+1) ?

The integral of (12x^2-4x+7)/(3x+1) is 2x^2-8/3 x+29/9 ln|3x+1|-2/3-4/9+C

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integral of (12x^2-4x+7)/(3x+1)

\int \frac{12 x ^{2} - 4 x + 7}{3 x + 1} d x

2 x^{2} - \frac{8}{3} x + \frac{29}{9} ln ∣ 3 x + 1 ∣ - \frac{2}{3} - \frac{4}{9} + C

Solution steps

\int \frac{12 x ^{2} - 4 x + 7}{3 x + 1} d x

Expand

\frac{12 x ^{2} - 4 x + 7}{3 x + 1} : \frac{12 x ^{2}}{3 x + 1} - \frac{4 x}{3 x + 1} + \frac{7}{3 x + 1}

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{12 x ^{2}}{3 x + 1} d x - \int \frac{4 x}{3 x + 1} d x + \int \frac{7}{3 x + 1} d x

\int \frac{12 x ^{2}}{3 x + 1} d x = \frac{2}{9} (9 x^{2} - 6 x + 2 ln ∣ 3 x + 1 ∣ - 3)

\int \frac{4 x}{3 x + 1} d x = \frac{4}{9} (3 x + 1 - ln ∣ 3 x + 1 ∣)

\int \frac{7}{3 x + 1} d x = \frac{7}{3} ln ∣ 3 x + 1 ∣

= \frac{2}{9} (9 x^{2} - 6 x + 2 ln ∣ 3 x + 1 ∣ - 3) - \frac{4}{9} (3 x + 1 - ln ∣ 3 x + 1 ∣) + \frac{7}{3} ln ∣ 3 x + 1 ∣

Simplify

\frac{2}{9} (9 x^{2} - 6 x + 2 ln ∣ 3 x + 1 ∣ - 3) - \frac{4}{9} (3 x + 1 - ln ∣ 3 x + 1 ∣) + \frac{7}{3} ln ∣ 3 x + 1 ∣ : 2 x^{2} - \frac{8}{3} x + \frac{29}{9} ln ∣ 3 x + 1 ∣ - \frac{2}{3} - \frac{4}{9}

= 2 x^{2} - \frac{8}{3} x + \frac{29}{9} ln ∣ 3 x + 1 ∣ - \frac{2}{3} - \frac{4}{9}

Add a constant to the solution

= 2 x^{2} - \frac{8}{3} x + \frac{29}{9} ln ∣ 3 x + 1 ∣ - \frac{2}{3} - \frac{4}{9} + C

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\int x (2^{x}) d x

What is the integral of (12x^2-4x+7)/(3x+1) ?
The integral of (12x^2-4x+7)/(3x+1) is 2x^2-8/3 x+29/9 ln|3x+1|-2/3-4/9+C