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

The integral of (2x^3+1)/(x^2-4x) is x^2+8x-1/4 ln|x|+129/4 ln|x-4|-48+C

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

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

x^{2} + 8 x - \frac{1}{4} ln ∣ x ∣ + \frac{129}{4} ln ∣ x - 4 ∣ - 48 + C

Solution steps

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

Expand

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

\int \frac{2 x ^{3}}{x ^{2} - 4 x} d x = x^{2} + 8 x + 32 ln ∣ x - 4 ∣ - 48

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

= x^{2} + 8 x + 32 ln ∣ x - 4 ∣ - 48 - \frac{1}{4} ln ∣ x ∣ + \frac{1}{4} ln ∣ x - 4 ∣

Simplify

x^{2} + 8 x + 32 ln ∣ x - 4 ∣ - 48 - \frac{1}{4} ln ∣ x ∣ + \frac{1}{4} ln ∣ x - 4 ∣ : x^{2} + 8 x - \frac{1}{4} ln ∣ x ∣ + \frac{129}{4} ln ∣ x - 4 ∣ - 48

= x^{2} + 8 x - \frac{1}{4} ln ∣ x ∣ + \frac{129}{4} ln ∣ x - 4 ∣ - 48

Add a constant to the solution

= x^{2} + 8 x - \frac{1}{4} ln ∣ x ∣ + \frac{129}{4} ln ∣ x - 4 ∣ - 48 + C

What is the integral of (2x^3+1)/(x^2-4x) ?
The integral of (2x^3+1)/(x^2-4x) is x^2+8x-1/4 ln|x|+129/4 ln|x-4|-48+C