What is the integral of (x^2+10x+28)/(3x-1) ?

The integral of (x^2+10x+28)/(3x-1) is 1/6 x^2+31/9 x+283/27 ln|3x-1|-1/18-10/9+C

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integral of (x^2+10x+28)/(3x-1)

\int \frac{x ^{2} + 10 x + 28}{3 x - 1} d x

\frac{1}{6} x^{2} + \frac{31}{9} x + \frac{283}{27} ln ∣ 3 x - 1 ∣ - \frac{1}{18} - \frac{10}{9} + C

Solution steps

\int \frac{x ^{2} + 10 x + 28}{3 x - 1} d x

Expand

\frac{x ^{2} + 10 x + 28}{3 x - 1} : \frac{x ^{2}}{3 x - 1} + \frac{10 x}{3 x - 1} + \frac{28}{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{x ^{2}}{3 x - 1} d x + \int \frac{10 x}{3 x - 1} d x + \int \frac{28}{3 x - 1} d x

\int \frac{x ^{2}}{3 x - 1} d x = \frac{1}{54} (9 x^{2} + 6 x + 2 ln ∣ 3 x - 1 ∣ - 3)

\int \frac{10 x}{3 x - 1} d x = \frac{10}{9} (3 x - 1 + ln ∣ 3 x - 1 ∣)

\int \frac{28}{3 x - 1} d x = \frac{28}{3} ln ∣ 3 x - 1 ∣

= \frac{1}{54} (9 x^{2} + 6 x + 2 ln ∣ 3 x - 1 ∣ - 3) + \frac{10}{9} (3 x - 1 + ln ∣ 3 x - 1 ∣) + \frac{28}{3} ln ∣ 3 x - 1 ∣

Simplify

\frac{1}{54} (9 x^{2} + 6 x + 2 ln ∣ 3 x - 1 ∣ - 3) + \frac{10}{9} (3 x - 1 + ln ∣ 3 x - 1 ∣) + \frac{28}{3} ln ∣ 3 x - 1 ∣ : \frac{1}{6} x^{2} + \frac{31}{9} x + \frac{283}{27} ln ∣ 3 x - 1 ∣ - \frac{1}{18} - \frac{10}{9}

= \frac{1}{6} x^{2} + \frac{31}{9} x + \frac{283}{27} ln ∣ 3 x - 1 ∣ - \frac{1}{18} - \frac{10}{9}

Add a constant to the solution

= \frac{1}{6} x^{2} + \frac{31}{9} x + \frac{283}{27} ln ∣ 3 x - 1 ∣ - \frac{1}{18} - \frac{10}{9} + C

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What is the integral of (x^2+10x+28)/(3x-1) ?
The integral of (x^2+10x+28)/(3x-1) is 1/6 x^2+31/9 x+283/27 ln|3x-1|-1/18-10/9+C