What is the integral of (12x^3-3x^2)/(3x^4-x^3) ?

The integral of (12x^3-3x^2)/(3x^4-x^3) is ln|3x-1|+3ln(3|x|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (12x^3-3x^2)/(3x^4-x^3)

\int \frac{12 x ^{3} - 3 x ^{2}}{3 x ^{4} - x ^{3}} d x

ln ∣ 3 x - 1 ∣ + 3 ln (3 ∣ x ∣) + C

Solution steps

\int \frac{12 x ^{3} - 3 x ^{2}}{3 x ^{4} - x ^{3}} d x

Expand

\frac{12 x ^{3} - 3 x ^{2}}{3 x ^{4} - x ^{3}} : \frac{12 x ^{3}}{3 x ^{4} - x ^{3}} - \frac{3 x ^{2}}{3 x ^{4} - x ^{3}}

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

\int \frac{12 x ^{3}}{3 x ^{4} - x ^{3}} d x = 4 ln ∣ 3 x - 1 ∣

\int \frac{3 x ^{2}}{3 x ^{4} - x ^{3}} d x = 3 (ln ∣ 3 x - 1 ∣ - ln (3 ∣ x ∣))

= 4 ln ∣ 3 x - 1 ∣ - 3 (ln ∣ 3 x - 1 ∣ - ln (3 ∣ x ∣))

Simplify

4 ln ∣ 3 x - 1 ∣ - 3 (ln ∣ 3 x - 1 ∣ - ln (3 ∣ x ∣)) : ln ∣ 3 x - 1 ∣ + 3 ln (3 ∣ x ∣)

= ln ∣ 3 x - 1 ∣ + 3 ln (3 ∣ x ∣)

Add a constant to the solution

= ln ∣ 3 x - 1 ∣ + 3 ln (3 ∣ x ∣) + C

\int \frac{2 x ^{4} - x - 4}{x ^{3} + 8} d x

\int \frac{e ^{12 t}}{4 e ^{6 t} - 1} d t

\int \frac{6 x ^{2} - 1}{x} d x

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

\int \frac{2}{3} e^{- 2 t} d t

What is the integral of (12x^3-3x^2)/(3x^4-x^3) ?
The integral of (12x^3-3x^2)/(3x^4-x^3) is ln|3x-1|+3ln(3|x|)+C