What is the integral of (16x^2-48x+15)/(2x^3-7x^2+3x) ?

The integral of (16x^2-48x+15)/(2x^3-7x^2+3x) is 5ln|x|+2ln|2x-1|+ln|x-3|+C

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integral of (16x^2-48x+15)/(2x^3-7x^2+3x)

\int \frac{16 x ^{2} - 48 x + 15}{2 x ^{3} - 7 x ^{2} + 3 x} d x

5 ln ∣ x ∣ + 2 ln ∣ 2 x - 1 ∣ + ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{16 x ^{2} - 48 x + 15}{2 x ^{3} - 7 x ^{2} + 3 x} d x

Take the partial fraction of

\frac{16 x ^{2} - 48 x + 15}{2 x ^{3} - 7 x ^{2} + 3 x} : \frac{5}{x} + \frac{4}{2 x - 1} + \frac{1}{x - 3}

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

\int \frac{5}{x} d x = 5 ln ∣ x ∣

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

\int \frac{1}{x - 3} d x = ln ∣ x - 3 ∣

= 5 ln ∣ x ∣ + 2 ln ∣ 2 x - 1 ∣ + ln ∣ x - 3 ∣

Add a constant to the solution

= 5 ln ∣ x ∣ + 2 ln ∣ 2 x - 1 ∣ + ln ∣ x - 3 ∣ + C

e^{0}

\int z 1 - z d z

x \to 0 lim (\frac{x - 2}{∣ x - 2 ∣})

\int_{0}^{4} y - y + 2 d y

\frac{d}{d x} (4 x (x^{2} - 4))

What is the integral of (16x^2-48x+15)/(2x^3-7x^2+3x) ?
The integral of (16x^2-48x+15)/(2x^3-7x^2+3x) is 5ln|x|+2ln|2x-1|+ln|x-3|+C