What is the integral of (3x^3-7x^2+5x-1)/(3x-1) ?

The integral of (3x^3-7x^2+5x-1)/(3x-1) is (x^3)/3-x^2+x+C

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integral of (3x^3-7x^2+5x-1)/(3x-1)

\int \frac{3 x ^{3} - 7 x ^{2} + 5 x - 1}{3 x - 1} d x

\frac{x ^{3}}{3} - x^{2} + x + C

Solution steps

\int \frac{3 x ^{3} - 7 x ^{2} + 5 x - 1}{3 x - 1} d x

Simplify

\frac{3 x ^{3} - 7 x ^{2} + 5 x - 1}{3 x - 1} : (x - 1)^{2}

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

Expand

(x - 1)^{2} : x^{2} - 2 x + 1

= \int x^{2} - 2 x + 1 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 x^{2} d x - \int 2 x d x + \int 1 d x

\int x^{2} d x = \frac{x ^{3}}{3}

\int 2 x d x = x^{2}

\int 1 d x = x

= \frac{x ^{3}}{3} - x^{2} + x

Add a constant to the solution

= \frac{x ^{3}}{3} - x^{2} + x + C

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What is the integral of (3x^3-7x^2+5x-1)/(3x-1) ?
The integral of (3x^3-7x^2+5x-1)/(3x-1) is (x^3)/3-x^2+x+C