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

The integral of (3x^2-5x+1)/(x-4) is 19x+29ln|x-4|+3/2 (x-4)^2-76+C

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

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

19 x + 29 ln ∣ x - 4 ∣ + \frac{3}{2} (x - 4)^{2} - 76 + C

Solution steps

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

Expand

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

\int \frac{3 x ^{2}}{x - 4} d x = 3 (\frac{( x - 4 ) ^{2}}{2} + 8 (x - 4) + 16 ln ∣ x - 4 ∣)

\int \frac{5 x}{x - 4} d x = 5 (x - 4 + 4 ln ∣ x - 4 ∣)

\int \frac{1}{x - 4} d x = ln ∣ x - 4 ∣

= 3 (\frac{( x - 4 ) ^{2}}{2} + 8 (x - 4) + 16 ln ∣ x - 4 ∣) - 5 (x - 4 + 4 ln ∣ x - 4 ∣) + ln ∣ x - 4 ∣

Simplify

3 (\frac{( x - 4 ) ^{2}}{2} + 8 (x - 4) + 16 ln ∣ x - 4 ∣) - 5 (x - 4 + 4 ln ∣ x - 4 ∣) + ln ∣ x - 4 ∣ : 19 x + 29 ln ∣ x - 4 ∣ + \frac{3}{2} (x - 4)^{2} - 76

= 19 x + 29 ln ∣ x - 4 ∣ + \frac{3}{2} (x - 4)^{2} - 76

Add a constant to the solution

= 19 x + 29 ln ∣ x - 4 ∣ + \frac{3}{2} (x - 4)^{2} - 76 + C

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What is the integral of (3x^2-5x+1)/(x-4) ?
The integral of (3x^2-5x+1)/(x-4) is 19x+29ln|x-4|+3/2 (x-4)^2-76+C