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

The integral of (3x^2+1)/(x-4) is 3/2 (x-4)^2+24x-96+49ln|x-4|+C

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

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

\frac{3}{2} (x - 4)^{2} + 24 x - 96 + 49 ln ∣ x - 4 ∣ + C

Solution steps

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

Expand

\frac{3 x ^{2} + 1}{x - 4} : \frac{3 x ^{2}}{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{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{1}{x - 4} d x = ln ∣ x - 4 ∣

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

Simplify

3 (\frac{( x - 4 ) ^{2}}{2} + 8 (x - 4) + 16 ln ∣ x - 4 ∣) + ln ∣ x - 4 ∣ : \frac{3}{2} (x - 4)^{2} + 24 x - 96 + 49 ln ∣ x - 4 ∣

= \frac{3}{2} (x - 4)^{2} + 24 x - 96 + 49 ln ∣ x - 4 ∣

Add a constant to the solution

= \frac{3}{2} (x - 4)^{2} + 24 x - 96 + 49 ln ∣ x - 4 ∣ + C

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What is the integral of (3x^2+1)/(x-4) ?
The integral of (3x^2+1)/(x-4) is 3/2 (x-4)^2+24x-96+49ln|x-4|+C