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

The integral of (5x^3+4x^2+1)/(6x-4) is (5x^3)/(18)+(11x^2)/(18)+22/27 x+115/162 ln|6x-4|+C

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integral of (5x^3+4x^2+1)/(6x-4)

\int \frac{5 x ^{3} + 4 x ^{2} + 1}{6 x - 4} d x

\frac{5 x ^{3}}{18} + \frac{11 x ^{2}}{18} + \frac{22}{27} x + \frac{115}{162} ln ∣ 6 x - 4 ∣ + C

Solution steps

\int \frac{5 x ^{3} + 4 x ^{2} + 1}{6 x - 4} d x

Long division

\frac{5 x ^{3} + 4 x ^{2} + 1}{6 x - 4} : \frac{5 x ^{2}}{6} + \frac{11 x}{9} + \frac{22}{27} + \frac{115}{27 ( 6 x - 4 )}

= \int \frac{5 x ^{2}}{6} + \frac{11 x}{9} + \frac{22}{27} + \frac{115}{27 ( 6 x - 4 )} 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{5 x ^{2}}{6} d x + \int \frac{11 x}{9} d x + \int \frac{22}{27} d x + \int \frac{115}{27 ( 6 x - 4 )} d x

\int \frac{5 x ^{2}}{6} d x = \frac{5 x ^{3}}{18}

\int \frac{11 x}{9} d x = \frac{11 x ^{2}}{18}

\int \frac{22}{27} d x = \frac{22}{27} x

\int \frac{115}{27 ( 6 x - 4 )} d x = \frac{115}{162} ln ∣ 6 x - 4 ∣

= \frac{5 x ^{3}}{18} + \frac{11 x ^{2}}{18} + \frac{22}{27} x + \frac{115}{162} ln ∣ 6 x - 4 ∣

Add a constant to the solution

= \frac{5 x ^{3}}{18} + \frac{11 x ^{2}}{18} + \frac{22}{27} x + \frac{115}{162} ln ∣ 6 x - 4 ∣ + C

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What is the integral of (5x^3+4x^2+1)/(6x-4) ?
The integral of (5x^3+4x^2+1)/(6x-4) is (5x^3)/(18)+(11x^2)/(18)+22/27 x+115/162 ln|6x-4|+C