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

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

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

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

x - ln ∣ x ∣ + \frac{1}{2} ln ∣ 2 x + 1 ∣ + ln ∣ 2 x - 1 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

\int 1 d x = x

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

= x - ln ∣ x ∣ + \frac{1}{2} ln ∣ 2 x + 1 ∣ + ln ∣ 2 x - 1 ∣

Add a constant to the solution

= x - ln ∣ x ∣ + \frac{1}{2} ln ∣ 2 x + 1 ∣ + ln ∣ 2 x - 1 ∣ + C

(x^{3} + 6) e^{x}

\frac{d}{d x} (lo g_{3} (4 x - 3))

\int cot^{3} (\frac{x}{2}) csc^{4} (\frac{x}{2}) d x

f (x) = 5 e^{x}, at x = 4

\int (sin (2 x))^{3} d x

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