What is the integral of (3x^3-4x^2-14x+8)/(x^2-3x+1) ?

The integral of (3x^3-4x^2-14x+8)/(x^2-3x+1) is (3x^2)/2+5x-ln|4x^2-12x+4|+C

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integral of (3x^3-4x^2-14x+8)/(x^2-3x+1)

\int \frac{3 x ^{3} - 4 x ^{2} - 14 x + 8}{x ^{2} - 3 x + 1} d x

\frac{3 x ^{2}}{2} + 5 x - ln 4 x^{2} - 12 x + 4 + C

Solution steps

\int \frac{3 x ^{3} - 4 x ^{2} - 14 x + 8}{x ^{2} - 3 x + 1} d x

Long division

\frac{3 x ^{3} - 4 x ^{2} - 14 x + 8}{x ^{2} - 3 x + 1} : 3 x + 5 + \frac{- 2 x + 3}{x ^{2} - 3 x + 1}

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

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

\int 5 d x = 5 x

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

= \frac{3 x ^{2}}{2} + 5 x - ln 4 x^{2} - 12 x + 4

Add a constant to the solution

= \frac{3 x ^{2}}{2} + 5 x - ln 4 x^{2} - 12 x + 4 + C

f (x) = ∣ x - 9 ∣

arctan (2 t)

\frac{- u ^{2} + 5}{( u ^{2} + 5 ) ^{2}}

\int ln (9 x + 7) d x

f (x) = (2 x^{2} - 3)^{2}

What is the integral of (3x^3-4x^2-14x+8)/(x^2-3x+1) ?
The integral of (3x^3-4x^2-14x+8)/(x^2-3x+1) is (3x^2)/2+5x-ln|4x^2-12x+4|+C