What is the integral of (6x^2-32x+41)/(x-4) ?

The integral of (6x^2-32x+41)/(x-4) is 3x^2-8x+9ln|x-4|-16+C

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integral of (6x^2-32x+41)/(x-4)

\int \frac{6 x ^{2} - 32 x + 41}{x - 4} d x

3 x^{2} - 8 x + 9 ln ∣ x - 4 ∣ - 16 + C

Solution steps

\int \frac{6 x ^{2} - 32 x + 41}{x - 4} d x

Expand

\frac{6 x ^{2} - 32 x + 41}{x - 4} : \frac{6 x ^{2}}{x - 4} - \frac{32 x}{x - 4} + \frac{41}{x - 4}

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{6 x ^{2}}{x - 4} d x - \int \frac{32 x}{x - 4} d x + \int \frac{41}{x - 4} d x

\int \frac{6 x ^{2}}{x - 4} d x = 6 (\frac{( x - 4 ) ^{2}}{2} + 8 (x - 4) + 16 ln ∣ x - 4 ∣)

\int \frac{32 x}{x - 4} d x = 32 (x - 4 + 4 ln ∣ x - 4 ∣)

\int \frac{41}{x - 4} d x = 41 ln ∣ x - 4 ∣

= 6 (\frac{( x - 4 ) ^{2}}{2} + 8 (x - 4) + 16 ln ∣ x - 4 ∣) - 32 (x - 4 + 4 ln ∣ x - 4 ∣) + 41 ln ∣ x - 4 ∣

Simplify

6 (\frac{( x - 4 ) ^{2}}{2} + 8 (x - 4) + 16 ln ∣ x - 4 ∣) - 32 (x - 4 + 4 ln ∣ x - 4 ∣) + 41 ln ∣ x - 4 ∣ : 3 x^{2} - 8 x + 9 ln ∣ x - 4 ∣ - 16

= 3 x^{2} - 8 x + 9 ln ∣ x - 4 ∣ - 16

Add a constant to the solution

= 3 x^{2} - 8 x + 9 ln ∣ x - 4 ∣ - 16 + C

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What is the integral of (6x^2-32x+41)/(x-4) ?
The integral of (6x^2-32x+41)/(x-4) is 3x^2-8x+9ln|x-4|-16+C